Practice
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Practice session
Catalog view: 759 problems available.
From AC and DC
Zero-mean sine RMS. An ideal sine voltage has 0 V DC offset and 2.00 V amplitude, so it ranges from −2.00 to +2.00 V. What is its total RMS over one complete cycle? Give four significant figures.
Show worked solution
For a sine with no offset, VRMS = A/√2 = 2/√2 = 1.414 V. Its signed mean is zero, but its RMS is not.
From AC and DC
Total RMS with DC offset. A waveform is v(t) = 1.00 + 2.00 sin(2πft) volts. Its AC-component RMS is 2/√2 V. What is the total RMS including the 1.00 V offset?
Show worked solution
VRMS,total = √(1² + (2/√2)²) = √3 = 1.732 V. Adding offset and amplitude directly would not calculate RMS.
From AC and DC
Peak-to-peak voltage. An ideal sine has amplitude 3.00 V and a DC offset of −1.00 V. What is its peak-to-peak voltage?
Show worked solution
Peak-to-peak is twice the sine amplitude: 2 × 3.00 = 6.00 V. The offset shifts both extrema equally and does not change their separation.
From AC and DC
Period from frequency. An ideal sine completes 50 cycles each second. What is its period in milliseconds?
Show worked solution
T = 1/f = 1/50 s = 0.020 s = 20.0 ms.
From AC and DC
Average ideal-resistor power. The waveform v(t) = 1.00 + 2.00 sin(2πft) V has total RMS √3 V. What average power does it produce in one ideal fixed 100 Ω resistor?
Show worked solution
Paverage = VRMS²/R = (√3)²/100 = 0.0300 W. This is average resistor power, not “RMS power.”
From Additive Manufacturing Process Families
Identify material extrusion
A machine softens thermoplastic filament, pushes it through a nozzle and deposits roads that bond as they cool. Which standard AM family describes the mechanism?
Show reasoning
Choice 2. Material is selectively dispensed through a nozzle. FDM is a registered trademark of Stratasys; it is not the generic family name.
From Additive Manufacturing Process Families
Cure selected regions in a vat
A build platform sits with liquid photopolymer. Light exposes selected regions for each layer, and the part later needs the specified cleaning and post-cure. Which family is described?
Show reasoning
Choice 1. The feedstock is liquid photopolymer in a vat and light selectively cures it. SLA and DLP are implementations with different exposure methods.
From Additive Manufacturing Process Families
Distinguish binder from fusion
A layer of metal powder is spread. A printhead deposits liquid only where a fragile green shape should form. The printed shape will later follow its specified debinding and sintering route. Which family is described?
Show reasoning
Choice 3. The powder is the build material and the jetted liquid is binder. The green shape is an intermediate state; later consolidation is part of this stated metal route.
From Additive Manufacturing Process Families
Follow the jetted material
A printhead deposits photopolymer droplets that become the part after UV curing. A second deposited material supports the part and is removed later. What distinguishes this material-jetting example from binder jetting?
Show reasoning
Choice 1. In this example the jetted photopolymer is build material. PolyJet is a Stratasys example within material jetting, not the generic family name.
From Additive Manufacturing Process Families
Classify an agent-assisted powder route
Polymer powder is spread. A printhead places fusing and detailing agents, then heat fuses selected regions. Loose powder surrounds the parts. Which family contains this Multi Jet Fusion example?
Show reasoning
Choice 2. The joining mechanism is thermal fusion of powder. A printhead alone does not determine the family; follow what the agent does to the build material.
From Additive Manufacturing Process Families
Decide whether printing is the last step
A specified metal binder-jet route ends the machine cycle with a depowdered green part. The process plan calls for debinding, sintering, dimensional inspection and machining of one mating face. Which statement is accurate?
Show reasoning
Choice 3. A green shape is not automatically the final dense metal part. The required route continues through its specified post-build steps, including inspection after dimensional change.
From Powder Bed Fusion in Polymers and Metals
Plan powder removal from a cavity
A polymer PBF part is drawn as a completely sealed hollow shell. Assume loose powder surrounds and fills the shell during the build, and the finished part must contain no loose powder. What is the best design response before building?
Show reasoning
Choice 2. Powder that supports geometry during the build can remain trapped afterward. The geometry needs an accessible exit and a way to verify the stated cleanup requirement.
From Powder Bed Fusion in Polymers and Metals
Reason about a metal PBF support
A specified laser-PBF metal process has a broad overhang prone to thermal distortion. The process engineer calls for supports that connect it to the build plate. Why can those supports matter in this stated case even though loose powder surrounds the overhang?
Show reasoning
Choice 3. In the stated laser-metal case, anchoring and heat conduction address thermal distortion. The answer is not a universal support rule for every PBF process.
From Powder Bed Fusion in Polymers and Metals
Plan post-build acceptance
A metal PBF machine completes a build with no alarm. The drawing specifies critical dimensions and a qualified material condition. The approved route also calls for cooldown, depowdering, support removal, heat treatment, machining and inspection. What is the accurate conclusion when the machine stops?
Show reasoning
Choice 1. Machine completion does not replace the route's required postprocessing or inspection. Acceptance depends on evidence against the stated drawing and material requirements.
Identify a radial load. A force acts vertically downward on a horizontal shaft. The shaft axis is horizontal. Which load is this?
Show explanation
Answer 1. The force is perpendicular to the shaft axis, so it is radial.
Recognize combined loading. A shaft carries a sideways belt pull and a separate force along its axis. What must its complete support arrangement carry?
Show explanation
Answer 2. The two perpendicular components make combined loading. They may be assigned to different bearings.
Select for repeated starts. A small shaft repeatedly starts and runs at moderate speed under radial load and modest thrust in either direction. Which is a sensible first catalog candidate if low starting resistance matters?
Show explanation
Answer 3. A deep-groove ball bearing can carry radial load and thrust in both directions with generally low starting resistance. Final selection still needs speed, load, life and lubrication checks.
Avoid the plain-bearing speed myth. A continuously running high-speed shaft already has an engineered oil supply. Must all plain bearings be rejected for speed alone?
Show explanation
Answer 2. Fluid-film bearings serve high-speed machinery. Their design and oil conditions matter; the category name is not a speed limit.
From Shaft Support and Axial Adjustment
Find the locating support. Support A retains both rings axially and is rated for radial load and thrust both ways. Support B carries radial load while allowing axial travel. Which end locates the shaft?
Show explanation
Answer 1. A provides axial location and carries the applied thrust. B accommodates relative axial growth.
From Shaft Support and Axial Adjustment
Predict exhausted floating travel. The shaft wants more thermal growth than the floating seat has travel. What happens in the ideal model?
Show explanation
Answer 3. Floating works only while sufficient travel remains. After the limit is reached, the support arrangement resists further expansion.
Count round-shaft freedoms. In the ideal geometric model, an adequately supported round sleeve has no anti-rotation key or axial stop. How many rigid-body freedoms remain? This is not a rotary service rating.
Show explanation
Answer 2. Translation and rotation about the shaft remain: two freedoms. Four are restrained. Ball-bushing rotary service still requires a suitable product rating.
Count profile-guide freedoms. A properly engaged profile carriage supports its rated transverse forces and moments. Which intended motion remains?
Show explanation
Answer 3. The profile geometry restrains the other five ideal rigid-body motions.
From Shaft Support and Axial Adjustment
Choose a rigid spacer. An assembly instruction specifies setting an end-cover position with a measured rigid spacer thickness. Which part performs that job?
Show explanation
Answer 1. A shim sets distance in the stack. The assembly geometry determines how its thickness changes end play.
From Shaft Support and Axial Adjustment
Maintain preload over small movement. An arrangement calls for a specified preload while allowing small dimensional changes. Which approach is appropriate?
Show explanation
Answer 2. A properly selected wave or Belleville spring applies force over a working deflection range. The required preload is a specification, not a feel test.
From Shaft Support and Axial Adjustment
Calculate free shaft growth. A 600 mm carbon-steel shaft warms uniformly by 40 °C and is free to expand. Use ΔL = α·L·ΔT, where L is the original length and ΔT is the temperature rise. For this estimate, use a constant α = 10.44 × 10−6 per °C. How much does the shaft length increase, in mm? Keep at least four significant figures in calculations and report four significant figures.
Show worked solution
The DOE Fundamentals Handbook, Material Science, Volume 2, Table 1 (MS-03, page 2) lists carbon steel at 5.8 × 10−6 per °F. A temperature rise of 1 °C equals a rise of 1.8 °F, so α = 5.8 × 10−6 × 1.8 = 10.44 × 10−6 per °C.
ΔL = (10.44 × 10−6 /°C)(600 mm)(40 °C) = 0.25056 mm ≈ 0.2506 mm. Using mm for L gives mm for ΔL. Four significant figures describe the arithmetic, not the accuracy of this approximate material coefficient.
Ball rating life. For an example ball bearing, C = 14 kN and P = 2 kN. Find L10 = (C/P)³ in million revolutions. Keep at least four significant figures in intermediate values and report four significant figures.
Show worked solution
(14/2)³ = 7³ = 343.0 million revolutions.
Life in hours. An example ball bearing has L10 = 343 million revolutions and runs at n = 1500 r/min. Find L10h = L10 × 10⁶/(60n). Keep at least four significant figures in intermediate values and report four significant figures.
Show worked solution
343 × 10⁶/(60 × 1500) = 3811.111… ≈ 3811 h.
Double the load. An example ball bearing has L10 = 343 million revolutions. Double P while keeping C fixed. Find the new L10 in million revolutions. Keep at least four significant figures in intermediate values and report four significant figures.
Show worked solution
The load is in the denominator and is cubed: 343/2³ = 42.875 ≈ 42.88 million revolutions.
Roller rating life. For an example roller bearing, C = 14 kN and P = 2 kN. Find L10 = (C/P)^(10/3) in million revolutions. Keep the exponent as 10/3. Keep at least four significant figures in intermediate values and report four significant figures.
Show worked solution
7^(10/3) = 656.1354… ≈ 656.1 million revolutions. Use 10/3, not 3, for the roller exponent.
Required dynamic rating. An example ball bearing must have L10 = 500 million revolutions at P = 2 kN. Find the required C = P × L10^(1/3), in kN. This is an exercise, not a catalog selection. Keep at least four significant figures in intermediate values and report four significant figures.
Show worked solution
C = 2 × 500^(1/3) = 15.87401… ≈ 15.87 kN. L10 enters as 500 million revolutions, not 500 × 10⁶.
From Belt, Chain and Gear Drives
1. Synchronize separated shafts: Two separated parallel shafts must stay synchronized in a clean area where oil at the drive contact is unwanted. Which ordinary drive is the best starting candidate? 1 = flat belt; 2 = timing belt; 3 = lubricated roller chain; 4 = a single compact gear pair.
Show worked solution
2 — timing belt. Teeth maintain timing and the belt contact runs dry. A flat belt can slip; ordinary chain needs joint lubrication; a compact gear pair cannot bridge the stated gap. This is a shortlist choice, followed by load and environment checks.
From Belt, Chain and Gear Drives
2. Quiet fan drive on a budget: A sheltered fan has separated parallel shafts. Low initial cost and quiet running matter, and a little speed variation is acceptable. Which is the best starting candidate? 1 = V-belt; 2 = precision helical gearbox with extra shafts; 3 = exposed roller chain; 4 = precision timing drive solely to guarantee synchronization.
Show worked solution
1 — V-belt. The job accepts the speed variation of a friction drive and benefits from a simple, quiet flexible span. The other choices add requirements or costs the scenario does not call for. Check the actual power and pulley ratings before selection.
From Chain Drives
3. Match an existing chain interface: A sheltered low-speed conveyor has serviceable ANSI #40 sprockets, suitable lubrication access and no requirement for quiet precision positioning. Which replacement transmission fits these specified wheels? 1 = flat belt; 2 = V-belt; 3 = compatible #40 roller chain; 4 = 08B chain chosen only because its pitch is also 12.7 mm.
Show worked solution
3 — compatible #40 roller chain. Match chain series and dimensions as well as pitch. Equal pitch alone does not make 08B interchangeable with #40. Confirm the strand count, load rating and connecting-link type too.
From Gear Drives and Tooth Geometry
4. Quieter drive in a compact housing: A compact lubricated gearbox has close parallel shafts and needs smoother high-speed engagement than a comparable spur pair. The bearing arrangement can carry axial thrust. Which candidate best addresses that requirement? 1 = a spur pair selected only for zero axial thrust; 2 = a slipping flat belt; 3 = a long exposed chain; 4 = a matched helical pair.
Show worked solution
4 — a matched helical pair. Progressive engagement generally improves smoothness and noise. The stated bearing capability addresses its axial thrust, but the tooth and bearing ratings still need checking.
From Chain Drives
5. Read #25 pitch: What is the nominal pitch of ANSI #25 chain, in inches? Enter a decimal to three places.
Show worked solution
Remove the final digit of 25: 2/8 = 0.250 in. The answer is 0.250 in. The final 5 describes bush-chain construction; it does not mean a 5 in pitch.
From Chain Drives
6. Read #35 pitch: What is the nominal pitch of ANSI #35 chain, in inches? Enter a decimal to three places.
Show worked solution
Remove the final digit: 3/8 = 0.375 in. The answer is 0.375 in, not 35/8 in and not 0.35 in.
From Chain Drives
7. Read #40-2 pitch: An ANSI #40-2 chain has two parallel strands. What is its nominal pitch in inches? Enter a decimal to three places.
Show worked solution
The base number is 40: 4/8 = 0.500 in. The answer is 0.500 in. The -2 changes the strand count, not the pitch.
From Chain Drives
8. Look up 08B pitch: A B-series lookup lists 05B-1 at 8.000 mm, 06B-1 at 9.525 mm, 08B-1 at 12.700 mm and 10B-1 at 15.875 mm. What pitch should a student record for 08B-1, to two decimal places in millimeters?
Show worked solution
The 08B-1 row gives 12.70 mm. For this size, 8/16 in is 0.5 in, and 0.5 × 25.4 = 12.70 mm. Do not read 08 as 8 mm, and do not generalize this fraction rule to small sizes such as 05B.
From Chain Drives
9. Write a sprocket reduction: A 22-tooth driver turns a 55-tooth driven sprocket with compatible chain and normal engagement. Write the reduction as X:1, input speed over output speed. Enter only X, to two decimal places.
Show worked solution
i = Ndriven/Ndriver = 55/22 = 2.50. The reduction is 2.50:1. Enter 2.50. Reversing the tooth counts gives 0.40, the reciprocal, not this reduction. See Gear Ratios.
From Chain Drives
10. Find driven-sprocket speed: A 21-tooth sprocket at 980 rpm drives a 49-tooth sprocket through correctly engaged chain. What is the average driven speed? Round to the nearest rpm; ignore chordal speed ripple.
Show worked solution
nout = 980 × 21/49 = 420 rpm. The answer is 420 rpm. A larger driven sprocket turns slower. The inverse calculation would give about 2287 rpm, which fails that direction check. See Gear Ratios.
From Belt Drives
11. Find timing-pulley speed: A timing belt connects a 28-tooth driver at 1440 rpm to a 60-tooth driven pulley. The belt and pulleys match, with no tooth jumping. What is the output speed, rounded to the nearest rpm?
Show worked solution
nout = 1440 × 28/60 = 672 rpm. The answer is 672 rpm. Teeth, not outside pulley diameters, establish this ratio. See Gear Ratios.
From Belt Drives
12. Estimate V-belt output speed: A V-belt driver has an effective diameter of 90 mm and runs at 1680 rpm. The driven sheave has an effective diameter of 210 mm. Estimate output speed to the nearest rpm, explicitly assuming no slip.
Show worked solution
nout = 1680 × 90/210 = 720 rpm. The no-slip estimate is 720 rpm. It is not a guarantee of actual friction-belt speed under load. Both dimensions use the same effective-diameter convention. See Gear Ratios.
From Belt Drives
Find an open-belt pitch length: An open belt has pitch diameters D = 200 mm and d = 100 mm, with C = 400 mm. Find the approximate pitch length L using the unequal-pulley correction. Round to 0.1 mm. You may use π = 3.14 and four-significant-figure intermediate values.
Show worked solution
L ≈ 2(400) + π(200 + 100)/2 + (200 − 100)²/(4 × 400) = 800 + 471.2389 + 6.25 = 1277.5 mm. Omitting the correction gives 1271.24 mm.
From Belt Drives
Find spacing for a chosen belt length: For a hypothetical stock choice with pitch length L = 1300 mm, use D = 200 mm and d = 100 mm. Find C from the inverse length approximation. This is a teaching value, not a catalog listing. Round to 0.1 mm. You may use π = 3.14 and four-significant-figure intermediate values.
Show worked solution
B = 4(1300) − 2π(200 + 100) = 3315.0444 mm. C = [B + √(B² − 32(100)²)]/16 ≈ 411.3 mm. Substitution into the forward equation returns 1300 mm.
From Belt Drives
Find large-pulley driven speed: The small driver has pitch diameter d = 100 mm and turns at 1200 r/min. The large driven pulley has D = 200 mm. Assuming no slip, find its speed to the nearest r/min. You may use π = 3.14 and four-significant-figure intermediate values.
Show worked solution
n_small/n_large = D/d = 2. Therefore n_large = 1200/2 = 600 r/min. Multiplying by 2 gives 2400 r/min and reverses the reduction.
From Belt Drives
Find belt linear speed: A small pulley with pitch diameter d = 100 mm turns at 1200 r/min without slip. Find belt speed in m/min. Round to 0.1 m/min. You may use π = 3.14 and four-significant-figure intermediate values.
Show worked solution
v = πd n_small / 1000 = π × 100 × 1200 / 1000 = 377.0 m/min. Use diameter, not the 50 mm radius. Using π = 3.14 gives 376.8 m/min.
From Belt Drives
Find small-pulley wrap: An open belt uses D = 200 mm, d = 100 mm and C = 400 mm. Find the small-pulley wrap θ = 180° − 2 asin((D − d)/(2C)). Express the inverse sine in degrees. Round to 0.1°. You may use π = 3.14 and four-significant-figure intermediate values.
Show worked solution
α = asin(100/800) = 7.180756°. θ = 180° − 2α = 165.6°. Subtracting the raw radian value from 180 gives about 179.75°, which mixes units.
From Machining Keyseats and Keyways
A through hub keyway. A hub has a finished through bore and needs a straight internal keyway. A matching broach set and suitable press are available. Choose the direct method.
- End mill a slot on the outside of a shaft.
- Use a guided push keyway broach.
- Use a Woodruff cutter to make a curved shaft recess.
Show reasoning
Option 2. The bushing guides the broach inside the existing bore; the feature is a hub keyway.
From Machining Keyseats and Keyways
A keyseat ending within the shaft. A straight flat-bottom shaft keyseat ends before either shaft end. Rounded ends in plan view are allowed. Which tool suits this feature?
- An end mill with a permitted entry method.
- A conventional through-hole keyway broach and bore bushing.
- A drill alone.
Show reasoning
Option 1. An end mill can enter and travel along this closed-end slot. Its diameter sets the end radius.
From Broaching
Repeated through hex holes. Many identical parts need a through hex hole. A prepared starting hole, suitable press and dedicated hex hole broach are specified. Which method directly produces the profile?
- A round reamer alone.
- A Woodruff keyseat cutter.
- Conventional internal hole broaching.
Show reasoning
Option 3. A dedicated hole broach generates the polygon from its specified starting hole. A keyway broach cuts only a slot, not the entire hex.
From Broaching
Between keyway passes. A multi-pass keyway set calls for one added shim after its first pass. Choose the correct next sequence after that pass is complete.
- Add an arbitrary shim while the broach remains loaded.
- Remove and clean the broach; insert the specified shim behind it; reinsert, lubricate, check alignment and make the next pass.
- Repeat without a shim to obtain the specified extra depth.
Show reasoning
Option 2. The shim offsets the tool for the next cut. The prescribed stack belongs behind the broach, and setup checks repeat between passes.
From Broaching
Count first and shimmed passes. Invented planning model: the unshimmed pass reaches 1.0 mm radial depth. Each later pass, with one additional shim increment, adds 0.5 mm. How many total passes reach 2.5 mm? Include the first pass; round a fractional added-pass count upward. These are not tool settings.
Show reasoning
1 + ceiling((2.5 − 1.0)/0.5) = 1 + 3 = 4 passes. Depths are 1.0, 1.5, 2.0 and 2.5 mm. Three counts only the added passes; five incorrectly divides the full target by the shim increment.
From Broaching
Round up, then check overshoot. Invented model: the first pass reaches 1.0 mm; each added shim increment gives 0.5 mm more depth on the next pass. What minimum total pass count reaches AT LEAST 2.2 mm? Round upward to a whole pass. This counting answer is not permission to exceed a drawing limit.
Show reasoning
1 + ceiling((2.2 − 1.0)/0.5) = 1 + ceiling(2.4) = 4. Three passes reach only 2.0 mm. Four reach 2.5 mm, so this schedule is unsuitable if 2.2 mm is the maximum allowed depth.
From Keys and Keyways in Shaft-Hub Connections
Equal cross-section dimensions. A parallel key has equal width and height and a uniform section along its working length. Identify it.
- Square parallel key.
- Rectangular parallel key with unequal width and height.
- Woodruff key.
Show reasoning
Option 1. Square describes equal width and height in cross-section, not the shape of the key seen from above.
From Keys and Keyways in Shaft-Hub Connections
Unequal cross-section dimensions. A parallel key has unequal width and height, with no taper along its working length. Identify it.
- Woodruff key.
- Rectangular parallel key.
- Gib-head key.
Show reasoning
Option 2. This is the rectangular parallel form; the dimensions alone do not specify the required fit.
From Keys and Keyways in Shaft-Hub Connections
A disk segment seated in a shaft. A key has a curved bottom that sits in a circular recess cut into the shaft. Identify the key.
- Square parallel key.
- Gib-head key.
- Woodruff key.
Show reasoning
Option 3. The Woodruff key is a disk segment with a matching curved shaft seat.
From Keys and Keyways in Shaft-Hub Connections
A tapered key with a removal head. A tapered key has a projecting head that can be engaged to remove it. Identify the key.
- Gib-head key.
- Woodruff key.
- Headless rectangular parallel key.
Show reasoning
Option 1. The gib head supplies a place to engage the key for removal.
Recover a hinge's intended spin
In the ideal axis-aligned lab, a grounded housing supports a shaft/lever. Its shoulder is seated at z = 0 and angular alignment holds θ = 0°. The goal is a hinge that turns without axial slide. Which change meets that goal?
Show answer and reasoning
Remove angular alignment; keep the shoulder seated. The hinge needs one rotation and no axial translation. Removing only the angular condition restores spin; z = 0 still blocks slide. More constraints are not automatically better for a moving joint.
Preserve a slider's axial travel
In the same bounded two-coordinate lab, axis alignment is already built in. The goal is axial sliding with no spin. Which additional condition matches it? Ignore contact, friction and collision; travel is within the declared lab range.
Show answer and reasoning
Angular alignment only. It blocks spin and leaves z free to vary. Shoulder seating would remove the intended slide. One DOF alone does not identify a joint: a hinge and a slider each have one, but preserve different motions.
Apply a locating datum
The lab shaft is at z = 18 mm and θ = 70° with neither additional condition active. Select shoulder seating, explicitly defined as z = 0. What is the resulting z coordinate in millimeters? This condition leaves spin free.
Show answer and reasoning
0 mm. The condition locates the shoulder at its stated datum, rather than freezing the existing 18 mm position. The angle stays 70° until another motion or condition changes it.
Count a free component’s motions
A rigid component floats freely in a three-dimensional assembly. It has no mates, contacts, or fixed state. How many independent degrees of freedom does it have relative to the assembly frame?
Show answer and reasoning
6 DOF. Three translations and three rotations: 3 + 3 = 6. Three is the count of axes, not the count of independent rigid-body motions.
Seat a flat plate on a base
A free rigid plate receives one coincident mate between its flat underside and the top plane of a grounded base. No other mates or contact constraints exist. How many DOF remain?
Show answer and reasoning
3 DOF. The plate can slide in two directions within the plane and rotate about the plane normal. The mate removes normal translation and two tilting motions: 6 − 3 = 3.
Align a shaft with a bore
A free shaft is mated concentrically to the cylindrical bore of a grounded housing. Rotation is not locked. There are no other mates or contacts. How many DOF remain?
Show answer and reasoning
2 DOF. Translation along the common axis and rotation about that axis remain. The axis alignment removes four of the initial six independent motions: 6 − 4 = 2.
Seat the shaft shoulder
A shaft already has a concentric mate to a grounded bore, with rotation unlocked. Add a coincident mate between its shoulder face and a housing end face, both perpendicular to the common axis. No other constraints exist. How many DOF remain?
Show answer and reasoning
1 DOF. Concentric alignment left axial slide and spin. Seating the shoulder removes the slide. Rotation about the common axis remains: 2 − 1 = 1.
Lock the remaining rotation
A shaft’s axis and shoulder are located as in the previous type of joint: only rotation about its axis remains. An independent angular constraint now fixes that remaining rotation. How many DOF remain? Assume compatible constraints.
Show answer and reasoning
0 DOF. The final independent rotation is removed: 1 − 1 = 0. This is appropriate for a stationary part; a pivot meant to turn should retain that rotation.
From Exploded Views and Bills of Materials
Count component instances
| Item | Part number | Description | Qty per assembly |
|---|---|---|---|
| 1 | PV-110 | Support | 1 |
| 2 | PV-210 | Lever | 1 |
| 3 | PV-310 | Pivot pin | 1 |
| 4 | HW-410 | Thrust washer | 2 |
| 5 | HW-510 | Pin retainer | 1 |
How many individual component instances are required for one assembly? Count both washers.
Show answer and reasoning
6 components. Add the quantity column: 1 + 1 + 1 + 2 + 1 = 6. Five is the number of distinct rows, not the number of instances.
From Exploded Views and Bills of Materials
Count washers for six assemblies
| Item | Part number | Description | Qty per assembly |
|---|---|---|---|
| 1 | PV-110 | Support | 1 |
| 2 | PV-210 | Lever | 1 |
| 3 | PV-310 | Pivot pin | 1 |
| 4 | HW-410 | Thrust washer | 2 |
| 5 | HW-510 | Pin retainer | 1 |
How many HW-410 washers are required for six complete assemblies, with no spares?
Show answer and reasoning
12 washers. 2 washers/assembly × 6 assemblies = 12 washers. Item number 4 identifies the row; using 4 × 6 = 24 mistakes an identifier for a quantity.
Choose mates that fully locate a cover
A rigid rectangular cover must be stationary relative to a grounded rectangular housing. Their corresponding bottom, side, and end datum planes are mutually perpendicular and geometrically compatible. Choose A, B, or C and explain.
- A: Coincident bottom planes only.
- B: Coincident bottom planes, then a parallel mate between those same planes.
- C: Coincident bottom, corresponding side, and corresponding end planes.
Self-check: write your reasoning before revealing the explanation. This text is not automatically graded or saved.
Show answer and reasoning
C. Seating the bottom leaves two in-plane slides and one spin. The independent side-plane condition removes one slide and that spin. The end-plane condition removes the last slide. A leaves three DOF. B repeats an orientation already established by coincidence. This answer assumes matching perpendicular datum planes, not arbitrary selections.
Find the redundant relationship
A shaft has a concentric mate to a grounded housing bore, plus a seated shoulder mate. Which proposed addition repeats an existing geometric condition: A, a parallel relation between those same two axes; or B, an independent constraint fixing the remaining shaft rotation? Explain.
Self-check: write your reasoning before revealing the explanation. This text is not automatically graded or saved.
Show answer and reasoning
A is redundant. Coincident axes are already parallel. B removes the remaining spin; whether it is wanted depends on whether the shaft is meant to rotate. Do not add B merely to remove an under-defined indicator.
From Exploded Views and Bills of Materials
Read the washer balloon
| Item | Part number | Description | Qty per assembly |
|---|---|---|---|
| 1 | PV-110 | Support | 1 |
| 2 | PV-210 | Lever | 1 |
| 3 | PV-310 | Pivot pin | 1 |
| 4 | HW-410 | Thrust washer | 2 |
| 5 | HW-510 | Pin retainer | 1 |
A leader ends at a washer and its balloon reads 4. State the part number and quantity per assembly. Does seeing balloon 4 twice double the listed quantity?
Self-check: write your reasoning before revealing the explanation. This text is not automatically graded or saved.
Show answer and reasoning
HW-410; two per assembly. The balloon maps to item 4. Two leaders identify two instances of the same part; they do not multiply the quantity column again.
Choose a meaningful motion check
A lever clears its housing at both travel endpoints. A mate animation moves, but you have not checked intermediate positions or collision settings. Can you conclude that it clears the housing throughout its travel? Describe the next check.
Self-check: write your reasoning before revealing the explanation. This text is not automatically graded or saved.
Show answer and reasoning
No. Inspect intermediate positions across the intended range and use appropriate interference or collision checks on the whole connected assembly. Endpoint clearance and a moving preview do not establish clearance everywhere. Check the scope of the tool: a joint preview may ignore other mates, and ordinary dragging may allow solids to pass through each other.
Recover a reference after renaming
A part was renamed in ordinary File Explorer. Its assembly now reports the old filename as missing. Choose the best response: A, suppress the missing component permanently; B, verify the intended file and repair the reference using the CAD workflow, then save and reopen affected documents; or C, insert an unrelated part with the old name.
Self-check: write your reasoning before revealing the explanation. This text is not automatically graded or saved.
Show answer and reasoning
B. Confirm the correct component identity before repairing its reference. Save the affected documents and check drawings as well as assemblies. Suppressing the component hides the symptom; substituting an unrelated file can make a model open while describing the wrong design.
From CAD Certification Preparation
Mass from volume and density. For this original general skill check, assume a uniform aluminum solid has volume 12.0 cm³ and density 2.70 g/cm³. Calculate its mass in grams, rounded to the nearest 0.1 g.
Show worked solution
m = density × volume = 2.70 g/cm³ × 12.0 cm³ = 32.4 g. Cubic centimeters cancel. Dividing volume by density gives about 4.44 and does not give units of mass.
From CAD Certification Preparation
Convert a cubic measurement. In this original general skill check, a solid's volume is 8,750 mm³. Express that same volume in cm³ to the nearest 0.01 cm³.
Show worked solution
1 cm³ = (10 mm)³ = 1,000 mm³. Therefore 8,750 ÷ 1,000 = 8.75 cm³. Dividing by 10 only, as for length, would give 875, which is wrong for volume.
From CAD Certification Preparation
Change only the displayed units. In this original general skill check, you change a CAD measurement display from millimeters to centimeters without editing geometry or material. What happens to the physical part's mass?
Show explanation
It stays the same. A display conversion describes the same geometry using different units. Resizing the geometry or changing density would be a different operation.
From Drawing Terminology Review
Read a radius before entering a diameter. This original general skill check describes a circle with radius 7.0 mm. A sketch tool asks for diameter. Enter the diameter in mm to the nearest 0.1 mm.
Show worked solution
d = 2r = 2 × 7.0 mm = 14.0 mm. Entering 7.0 as the diameter makes the circle half the intended diameter.
From Drawing Terminology Review
Recognize an enlarged detail. In this original general skill check, a drawing enlarges a small region solely to make its features easier to read. Which term describes that view?
Show explanation
Detail view. Enlarging a selected region makes small features readable. A section reveals features at an imaginary cut; an auxiliary view establishes a viewing direction perpendicular to an inclined surface.
File references: A drawing shows a local SOLIDWORKS assembly made from separate part files. What should a transfer include?
- Only the drawing file.
- The drawing, assembly, and required component files.
- Only a screenshot of the assembly.
Show answer and reasoning
Choice 2. The drawing references its model, and the assembly references its components. Review a Pack and Go package rather than assuming one file holds everything.
Choosing an Onshape tab: You want to build two related parts from sketches and features in one Onshape modeling context. Where should you start?
- A Drawing tab.
- An Assembly tab that automatically creates their geometry.
- A Part Studio tab.
Show answer and reasoning
Choice 3. A Part Studio can create several related parts. An Assembly arranges instances and defines their relationships.
Changing displayed units: A correctly modeled edge is 1 inch long. You change its displayed units to millimeters without scaling the model. What length should it report?
- 25.4 mm.
- 1 mm.
- 254 mm.
Show answer and reasoning
Choice 1. One inch equals exactly 25.4 mm. A display-unit conversion preserves physical length.
From CAD Model Views and Checks
Normal to a plane: A planar face is tilted relative to all standard model planes. Which direction gives a normal view of that face?
- Along a direction perpendicular to the face.
- Always from the model front.
- Along an edge in the face.
Show answer and reasoning
Choice 1. Normal means perpendicular to the selected plane. It need not match a standard front, top, or side direction.
From CAD Model Views and Checks
Section view or cut feature: You turn on a section view in the model window, inspect a cavity, then turn the section off. What changed?
- Material was permanently removed.
- The display was clipped temporarily.
- A drawing section was automatically added.
Show answer and reasoning
Choice 2. The viewing section exposes the inside visually. It is not a material-removal feature or an automatically created drawing view.
From CAD Model Views and Checks
Checking a mass result: A uniform solid has volume 12 cm³ and assigned density 2.5 g/cm³. What mass should the model report?
- 4.8 g.
- 14.5 g.
- 30 g.
Show answer and reasoning
Choice 3. Mass = density × volume = 2.5 × 12 = 30 g. Dividing volume by density or adding the values gives the other choices, not mass.
From CAD Model Views and Checks
Interpreting a measurement: Measure reports the minimum gap between two cylindrical faces, but you need the distance between their axes. What should you do?
- Accept the number because every distance is equivalent.
- Check the selected entities and choose an appropriate center or axis measurement.
- Change the display color.
Show answer and reasoning
Choice 2. Minimum surface gap and axis-to-axis distance answer different questions. Read the measurement label and check the selection.
From CAD History, Versions, and Sharing
Version or branch: You need a fixed milestone and a separate place to explore a new design. Which Onshape workflow fits?
- Create a version, then branch from it to an editable workspace.
- Edit the captured design state of the version directly.
- Move the rollback bar and call that a separate branch.
Show answer and reasoning
Choice 1. A version preserves a design state. Branching from it creates a separate workspace for further changes.
From CAD History, Versions, and Sharing
Feature order or edit history: An early sketch was edited today. You move the rollback bar just after that sketch. Does this recover its dimensions from last week?
- Yes; feature position is a timestamp.
- Yes; rollback always restores the previous save.
- No; rollback controls feature evaluation order, not which historical edit is used.
Show answer and reasoning
Choice 3. The current sketch can contain today's edits even near the start of the feature sequence. Use history or a preserved earlier state to inspect older geometry.
From CAD History, Versions, and Sharing
Checking a shared link: A reviewer needs to inspect a fixed Onshape design milestone. What should you check before sending the link?
- Only that the document has a short name.
- That the link targets the intended version and the recipient has appropriate access.
- That the reviewer has edit permission to every workspace.
Show answer and reasoning
Choice 2. A fixed version identifies the milestone. Sharing permissions determine access; the URL alone does not grant it.
From Measuring Electric Circuits
Place probes for a positive R1 reading. A 12 V source supplies 30 Ω R1 and 60 Ω R2 in series, with both switches closed. Which ordered placement gives the positive drop across R1? 1 = red after S1, black after R1; 2 = red after R1, black after S1; 3 = both probes after S1; 4 = red supply −, black supply +.
Show worked solution
1. Current leaves the positive source terminal, passes S1, then crosses R1. Red before R1 and black after R1 gives V(red) − V(black) = +4 V.
From Measuring Electric Circuits
Reverse the probes across R1. The same closed 12 V, 30 Ω / 60 Ω series circuit has 4 V across R1. Put red after R1 and black after S1. What signed reading appears?
Show worked solution
V(red) − V(black) = 8 V − 12 V = −4 V. A negative sign gives polarity information; it is not an error.
From Measuring Electric Circuits
Measure across the only open switch. An ideal 12 V source supplies 30 Ω R1 and 60 Ω R2 in series. S1, immediately before R1, is open; S2, immediately before R2, is closed. Put red on supply + and black after S1. What signed voltage appears?
Show worked solution
No current flows, so neither resistor has a voltage drop. The downstream side is connected to supply − through the zero-current resistors. The sole open switch therefore has +12 V across it in this ideal circuit.
From Measuring Electric Circuits
Recognize a same-node reading. Both probes touch the same uninterrupted conductor after S1. What voltage does an ideal meter report?
Show worked solution
0 V. Both probes are at the same potential, whatever that potential is relative to supply −.
From Measuring Electric Circuits
Do not invent a floating-node voltage. Both series switches are open, isolating a conductor between them. What should this bounded ideal model report for a probe from that floating conductor to supply −? 1 = always 0 V; 2 = always 12 V; 3 = undetermined without another connection; 4 = 6 V because the resistors divide the source.
Show worked solution
3. Two breaks isolate a floating region. The one-open teaching model does not add leakage, capacitance, meter loading, or another path that would establish its potential.
From Absolute and Incremental Positioning
Point list: final X increment. In mm, start at A=(3,7), then visit B=(18,7), C=(18,−2), D=(8,−2). Convert the complete list to signed increments on paper. What is ΔX for C→D?
Show worked solution
The increments are A→B=(15,0), B→C=(0,−9), C→D=(−10,0), each found by next minus current. The requested value is −10 mm.
From Absolute and Incremental Positioning
Point list: middle Y increment. In mm, start at A=(3,7), then visit B=(18,7), C=(18,−2), D=(8,−2). Convert the complete list to signed increments on paper. What is ΔY for B→C?
Show worked solution
The increments are A→B=(15,0), B→C=(0,−9), C→D=(−10,0), each found by next minus current. The requested value is −9 mm.
From Absolute and Incremental Positioning
Increment list: final X. Start at (−4,6) mm. Apply increments (9,0), (0,−8), (−3,5). Write each absolute point. What is the final X?
Show worked solution
Successive points are (5,6), (5,−2), (2,3): add each increment to the previous point. Final X is 2 mm.
From Absolute and Incremental Positioning
Increment list: final Y. Start at (−4,6) mm. Apply increments (9,0), (0,−8), (−3,5). Write each absolute point. What is the final Y?
Show worked solution
Successive points are (5,6), (5,−2), (2,3): add each increment to the previous point. Final Y is 3 mm.
From Absolute and Incremental Positioning
Program endpoint: X. Paper trace only: start at (2,3,10) mm, no compensation, unchanged work offset. Common milling modes apply.
G21 G90 G94
G01 X14 Y9 F100
G91 X-4
Y6
G90 X5
M30Show worked solution
Positions after the four moves are (14,9,10), (10,9,10), (10,15,10), (5,15,10). M30 ends the trace without a positioning command. X is 5 mm.
From Absolute and Incremental Positioning
Program endpoint: Y. Paper trace only: start at (2,3,10) mm, no compensation, unchanged work offset. Common milling modes apply.
G21 G90 G94
G01 X14 Y9 F100
G91 X-4
Y6
G90 X5
M30Show worked solution
Positions after the four moves are (14,9,10), (10,9,10), (10,15,10), (5,15,10). M30 ends the trace without a positioning command. Y is 15 mm.
From Absolute and Incremental Positioning
Program endpoint: Z. Paper trace only: start at (2,3,10) mm, no compensation, unchanged work offset. Common milling modes apply.
G21 G90 G94
G01 X14 Y9 F100
G91 X-4
Y6
G90 X5
M30Show worked solution
Positions after the four moves are (14,9,10), (10,9,10), (10,15,10), (5,15,10). M30 ends the trace without a positioning command. No block moves Z; it remains 10 mm.
From G-code and Toolpaths
Quarter arc: I offset. In G17, G90, millimetres, with I/J measured from the start: a quarter-circle G03 goes from (5,7) to (13,15) around centre (5,15). What is I?
Show worked solution
I = centre X − start X = 5−5 = 0 mm. J = 15−7 = 8 mm. Both endpoint radii are 8 mm.
From G-code and Toolpaths
Quarter arc: J offset. In G17, G90, millimetres, with I/J measured from the start: a quarter-circle G03 goes from (5,7) to (13,15) around centre (5,15). What is J?
Show worked solution
J = centre Y − start Y = 15−7 = 8 mm. I = 0. Do not enter the absolute centre Y of 15.
From G-code and Toolpaths
Quarter arc: endpoint X. In the XY plane, start at (16,9) mm, centre (10,9), and move one quarter turn counterclockwise (G03). What is the absolute endpoint X?
Show worked solution
The start radius vector is (6,0). Rotating it 90° counterclockwise gives (0,6). The endpoint is (10,15); X is 10 mm.
From G-code and Toolpaths
Quarter arc: endpoint Y. In the XY plane, start at (16,9) mm, centre (10,9), and move one quarter turn counterclockwise (G03). What is the absolute endpoint Y?
Show worked solution
The radius is 6 mm. The endpoint is 6 mm above the centre, (10,15); Y is 15 mm.
From G-code and Toolpaths
Reverse arc: centre offset. A G02 quarter arc returns from (13,15) to (5,7) mm around centre (5,15). I/J are measured from this new start. What is I?
Show worked solution
I = 5−13 = −8 mm; J = 15−15 = 0. Reversing direction also changes the point from which centre offsets are measured.
From CNC Axes and Offsets
Diameter to radial distance. A lathe uses diameter programming in mm. At X36, how far is the programmed surface from the spindle centreline?
Show worked solution
Radial distance = diameter / 2 = 36/2 = 18 mm.
From CNC Axes and Offsets
Radius to diameter command. A surface is 11 mm from the spindle centreline on the positive tool side. What absolute X value represents it in diameter programming?
Show worked solution
X = 2 × radius = 2 × 11 = 22 mm. Radius mode would use X11.
From CNC Axes and Offsets
Diameter change versus tool travel. A diameter-mode X coordinate changes from X42 to X34 mm. What is the magnitude of radial inward tool travel?
Show worked solution
The diameter decreases by 42−34 = 8 mm, so radial travel is 8/2 = 4 mm inward.
From CNC Axes and Offsets
Work coordinate to machine coordinate. In a simplified XY translation with no tool offsets or other shifts, work zero is at machine (−80,−40) mm. A point is work (25,12). What is machine X?
Show worked solution
Machine point = (−80+25,−40+12) = (−55,−28). Machine X is −55 mm.
Identify M05. For the common milling meanings taught here, M05 means: 1 = stop spindle; 2 = end program; 3 = select work offset. Enter the choice number.
Show worked solution
1: stop spindle. M30 ends the program; G54 selects a work coordinate system. Check the actual control manual.
Identify G01. For common milling usage, G01 means: 1 = rapid positioning; 2 = straight-line feed motion; 3 = peck drilling. Enter the choice number.
Show worked solution
2: straight-line feed motion. G00 is rapid positioning; G83 commonly selects peck drilling.
Identify G54. For common milling usage, G54: 1 = stores a new zero automatically; 2 = homes the machine; 3 = selects a stored work coordinate system. Enter the choice number.
Show worked solution
3: selects a stored work coordinate system. Selection neither measures the work nor writes the offset.
Feed per revolution to feed per minute. Assume G21 G95, F0.2 mm/rev and a steady spindle speed of 700 rpm. What feed per minute does this represent?
Show worked solution
0.2 mm/rev × 700 rev/min = 140 mm/min. This is arithmetic, not a cutting recommendation.
From CNC Turning Spindle and Feed Modes
CSS RPM: inch-a. Assume a Fanuc-style lathe control, where G96 is constant surface speed (codes vary by control), and steady-state G96 at 240 SFM, diameter 2 in, and a clamp above the calculated RPM. What RPM is requested? Use π, not 3.82, and round to the nearest whole RPM.
Show worked solution
RPM = 240 × 12 / (π × 2) = 458.3662, or 458 RPM. The diameter, not the radius, belongs in the denominator.
From CNC Turning Spindle and Feed Modes
CSS RPM: inch-b. Assume a Fanuc-style lathe control, where G96 is constant surface speed (codes vary by control), and steady-state G96 at 150 SFM, diameter 0.75 in, and a clamp above the calculated RPM. What RPM is requested? Use π, not 3.82, and round to the nearest whole RPM.
Show worked solution
RPM = 150 × 12 / (π × 0.75) = 763.9437, or 764 RPM. The diameter, not the radius, belongs in the denominator.
From CNC Turning Spindle and Feed Modes
CSS RPM: metric-a. Assume a Fanuc-style lathe control, where G96 is constant surface speed (codes vary by control), and steady-state G96 at 120 m/min, diameter 40 mm, and a clamp above the calculated RPM. What RPM is requested? Use π, not 3.82, and round to the nearest whole RPM.
Show worked solution
RPM = 120 × 1000 / (π × 40) = 954.9297, or 955 RPM. The diameter, not the radius, belongs in the denominator.
From CNC Turning Spindle and Feed Modes
CSS RPM: metric-b. Assume a Fanuc-style lathe control, where G96 is constant surface speed (codes vary by control), and steady-state G96 at 80 m/min, diameter 16 mm, and a clamp above the calculated RPM. What RPM is requested? Use π, not 3.82, and round to the nearest whole RPM.
Show worked solution
RPM = 80 × 1000 / (π × 16) = 1591.5494, or 1592 RPM. The diameter, not the radius, belongs in the denominator.
From CNC Turning Spindle and Feed Modes
CSS with a clamp. In metric CSS, S100 requests 100 m/min at a 10 mm diameter. The RPM ceiling is 2000. Ignoring acceleration and other limits, what RPM is commanded? Round to a whole RPM.
Show worked solution
The unrestricted calculation is 100 × 1000 / (π × 10) = 3183.10 RPM. The lower of this and the ceiling is 2000 RPM.
From CNC Turning Spindle and Feed Modes
Why the clamp matters. Assume a Fanuc-style lathe control, where G96 is constant surface speed (codes vary by control). During facing under G96, why must the RPM ceiling be set before CSS? Choose 1: requested RPM falls toward zero; 2: requested RPM rises as diameter decreases; 3: G96 fixes RPM regardless of diameter. Enter the choice number.
Show worked solution
2. RPM is inversely proportional to diameter. The clamp limits that request. Below the crossover diameter, surface speed falls as diameter decreases.
From CNC Turning Setup and Offsets
Diameter-mode X command. The required OD is 32 mm. In diameter programming with correctly set offsets, what absolute X value represents that cylindrical surface? Enter the number only.
Show worked solution
X32. The tool reference is 16 mm radially from the spindle centreline, but X reports diameter.
From CNC Turning Setup and Offsets
Radius-mode X command. The required OD is 32 mm. In radius programming with correctly set offsets, what absolute X value represents that cylindrical surface? Enter the number only.
Show worked solution
X = 32 / 2 = 16 mm. Z is unaffected by this convention.
From CNC Turning Setup and Offsets
Diameter after an X change. A tool starts at X30 mm in diameter mode. Its X coordinate decreases by 4 mm, then it turns a cylindrical surface along Z. With correct offsets and no deflection, what diameter results?
Show worked solution
Final X = 30 − 4 = 26 mm diameter. The radial inward travel was 2 mm, not 4 mm.
From CNC Turning Spindle and Feed Modes
Feed per revolution to IPM. Assume a Fanuc-style lathe control, where G99 is feed per revolution (codes vary by control), inch units, G99, F0.007 in/rev and a steady 900 RPM. What feed per minute results? Round to 0.1 in/min.
Show worked solution
0.007 × 900 = 6.3 in/min. Multiply advance per revolution by revolutions per minute.
From CNC Turning Spindle and Feed Modes
Metric feed per minute. Assume a Fanuc-style lathe control, where G99 is feed per revolution (codes vary by control), metric units, G99, F0.18 mm/rev and a steady 750 RPM. What feed per minute results? Round to a whole mm/min.
Show worked solution
0.18 × 750 = 135 mm/min.
From CNC Turning Spindle and Feed Modes
Find the missing speed limit. Paper review only; this Fanuc-style lathe has no verified RPM ceiling set. The snippet is G21 G99, then G96 S100 M03, then G01 X0 F0.10. Which missing setting must be resolved before CSS? 1: an appropriate maximum-RPM clamp; 2: mill G17; 3: G98 to hold feed per revolution. Enter the choice number.
Show worked solution
1. Establish the setup-appropriate RPM ceiling before G96, using G50 S… on this assumed control. G98 is feed per minute here. This identifies one error; it does not approve the rest of the snippet.
From CNC Turning Program Walkthrough
Find the wrong feed mode. Paper review only, using this page’s Fanuc-style lathe meanings. Intended feed is 0.12 mm/rev. The snippet is G21 G98, G97 S800 M03, G01 Z-8 F0.12. What change matches the intended feed? 1: change G21 to G20; 2: change G98 to G99; 3: change G97 to G96. Enter the choice number.
Show worked solution
2. G98 interprets F0.12 as 0.12 mm/min. G99 interprets it as 0.12 mm/rev, giving 0.12 × 800 = 96 mm/min. The remaining snippet still needs a complete setup and prove-out.
From Coolant Concentration and Mixing
Apply a factor of 1.7: A hypothetical fluid reads 2.6 Brix and its supplied factor is 1.7. Calculate concentration to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice.
Show worked solution
2.6 × 1.7 = 4.42%. Using 2.6% forgets the factor; dividing by 1.7 uses it backward.
From Coolant Concentration and Mixing
Apply a factor of 1.25: A hypothetical fluid reads 3.8 Brix and its supplied factor is 1.25. Calculate concentration to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice.
Show worked solution
3.8 × 1.25 = 4.75%. Brix is the instrument reading, not automatically percent.
From Coolant Concentration and Mixing
Apply a factor of 1.6: A hypothetical fluid reads 4.4 Brix and its supplied factor is 1.6. Calculate concentration to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice.
Show worked solution
4.4 × 1.6 = 7.04%. Do not divide the reading by the factor.
From Coolant Concentration and Mixing
Is the tank lean or rich?: A hypothetical tank reads 3.2 Brix; the supplied factor is 1.5 and target is 5.5%. Calculate measured concentration minus target to the nearest 0.01 percentage point. A negative result means lean; positive means rich. Decide which before revealing the solution.
All values are hypothetical; use the actual product data sheet in practice.
Show worked solution
Measured concentration = 3.2 × 1.5 = 4.8%. Difference = 4.8 − 5.5 = −0.70 percentage points: lean relative to this assumed target. A real decision also checks the permitted range.
From Coolant Concentration and Mixing
Compare measured concentration with the target: A hypothetical tank reads 4.2 Brix; the supplied factor is 1.8 and target is 6.0%. Calculate measured concentration minus target to the nearest 0.01 percentage point. A negative result means lean; positive means rich. Decide which before revealing the solution.
All values are hypothetical; use the actual product data sheet in practice.
Show worked solution
4.2 × 1.8 = 7.56%. Difference = 7.56 − 6.0 = +1.56 percentage points: rich. Comparing 4.2 directly with 6.0 would wrongly call it lean.
From Coolant Concentration and Mixing
Find the missing tank volume: A hypothetical tank contains 34 L of serviceable coolant. Its allowed final fill is 50 L. How much premix can be added? Give liters to the nearest 0.01 L.
All values are hypothetical; use the actual product data sheet in practice.
Show worked solution
A = 50 − 34 = 16.00 L. This determines volume only; measure concentration before choosing the premix strength.
From Coolant Concentration and Mixing
Restore a lean tank with premix: A hypothetical tank contains 34 L at 4.0%. Bring it to 50 L at an assumed target of 6.0%. What premix concentration is required? Give percent to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice. Assume additive liquid volumes, uniform mixing and no loss. Top-up examples assume serviceable fluid and supplier-approved premix.
Show worked solution
A = 50 − 34 = 16 L. P = (50 × 6 − 34 × 4) ÷ 16 = 10.25%. Check: (34 × 4 + 16 × 10.25) ÷ 50 = 6%. Premix at 6% would leave the tank lean.
From Coolant Concentration and Mixing
Restore a rich tank with weaker premix: A hypothetical tank contains 40 L at 7.0%. Bring it to 50 L at an assumed target of 6.0%. What premix concentration is required? Give percent to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice. Assume additive liquid volumes, uniform mixing and no loss. Top-up examples assume serviceable fluid and supplier-approved premix.
Show worked solution
A = 10 L. P = (50 × 6 − 40 × 7) ÷ 10 = 2.00%. Check: (280 + 20) ÷ 50 = 6%. Plain water would instead leave 5.6%.
From Coolant Concentration and Mixing
Find volume from an available premix: A hypothetical tank has 30 L at 4.0%. An approved teaching premix is 8.0%; the desired tank concentration is 5.0%. The allowed final volume is 45 L. How much premix reaches the target? Give liters to the nearest 0.01 L.
All values are hypothetical; use the actual product data sheet in practice. Assume additive liquid volumes, uniform mixing and no loss. Top-up examples assume serviceable fluid and supplier-approved premix.
Show worked solution
A = 30 × (5 − 4) ÷ (8 − 5) = 10.00 L. Final volume is 40 L, below the 45 L limit. Concentrate balance: (30 × 4 + 10 × 8) ÷ 40 = 5%.
From Coolant Concentration and Mixing
Measure concentrate for the top-up batch: Prepare 16 L of hypothetical 10.25% premix. How many liters of concentrate are needed? Give liters to the nearest 0.01 L.
All values are hypothetical; use the actual product data sheet in practice. Assume additive liquid volumes, uniform mixing and no loss. Top-up examples assume serviceable fluid and supplier-approved premix.
Show worked solution
Concentrate = 16 × 10.25 ÷ 100 = 1.64 L. Water = 16 − 1.64 = 14.36 L. Put water in first, then add concentrate while mixing.
From Coolant Concentration and Mixing
Convert a water-to-concentrate ratio: A hypothetical mixing instruction says 19 parts water to 1 part concentrate, by volume. What percentage of the final mix is concentrate? Give percent to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice. Assume additive liquid volumes, uniform mixing and no loss. Top-up examples assume serviceable fluid and supplier-approved premix.
Show worked solution
Total parts = 19 + 1 = 20. Concentrate percentage = 100 × 1 ÷ 20 = 5.00%. Using 100 ÷ 19 gives 5.26%, which uses water rather than the whole mixture as the denominator. Reversing the named ingredients gives 95%, also wrong.
From Coolant Concentration and Mixing
Keep all the parts in the denominator: A hypothetical mixing instruction says 31 parts water to 1 part concentrate, by volume. What percentage of the final mix is concentrate? Give percent to the nearest 0.01%.
All values are hypothetical; use the actual product data sheet in practice. Assume additive liquid volumes, uniform mixing and no loss. Top-up examples assume serviceable fluid and supplier-approved premix.
Show worked solution
Total parts = 32. Concentrate percentage = 100 ÷ 32 = 3.125%, or 3.13%. Using 100 ÷ 31 gives about 3.23%; reversing the named ingredients gives 96.875%. Neither is the concentrate fraction requested.
From 3D Printing from CAD
Orient a bar for tension
A straight printed bar will be pulled along its length. Compare printing it standing on its end with printing its length parallel to the bed. Assume the same material and a sound print. Which is the better starting orientation, and why?
Self-check: write a reason before revealing. This response is not automatically graded or saved.
Show reasoning
Start with the length parallel to the bed. Printing upright puts the tension across the interfaces between layers, where separation is often easier. Laying the bar down allows paths within layers to run along its length. Actual strength still depends on material, paths, and bonding; test a representative part.
From Laser Cutting and Engraving from CAD
Separate an outline from a filled mark
An approved sheet must yield a rectangular label with a filled dark symbol on its face. Which operation belongs on the outside boundary, and which belongs on the filled symbol? Should a thin score line automatically use through-cut settings?
Self-check: write a reason before revealing. This response is not automatically graded or saved.
Show reasoning
Use a vector cutting path for the outside boundary and raster engraving for the filled symbol. A vector score follows a path but is intended to mark the surface, so it needs an appropriate scoring setup rather than through-cut settings. Verify the assignments in the job preview.
From Laser Cutting and Engraving from CAD
Compensate an outside width
A rectangular cutout must finish 30.00 mm wide. Assume a constant 0.18 mm kerf centered on the path, with no automatic compensation. What path-to-path width produces that finished outside width? Give millimeters to the nearest 0.01 mm.
Show worked solution
Each path moves outward by 0.18 / 2 = 0.09 mm. Across both sides, 30.00 + 2 × 0.09 = 30.18 mm. Subtracting kerf would shrink the part further; adding only half the kerf would compensate only one edge.
From Laser Cutting and Engraving from CAD
Compensate an opening
A rectangular opening must finish 12.00 mm wide. Assume a constant 0.22 mm kerf centered on the path and no automatic compensation. What path-to-path width is needed? Give millimeters to the nearest 0.01 mm.
Show worked solution
The waste is inside the opening. Move each path inward by 0.22 / 2 = 0.11 mm. The path width is 12.00 − 0.22 = 11.78 mm. Cutting removes half the kerf beyond each path and brings the finished opening back to 12.00 mm.
From Choosing a Fabrication Process
Choose a process for two shapes
Choose an initial process for (A) a flat cover of constant thickness with surface lettering and (B) a one-piece rounded grip whose cross section changes along its length. Approved sheet and printable materials both meet the stated service requirements. Explain a limitation of each choice.
Self-check: write a reason before revealing. This response is not automatically graded or saved.
Show reasoning
A: laser-cut and engrave the sheet; account for kerf, available thickness, and edge quality. B: print the grip; choose orientation for load and surface quality and check support access. A laser could make a grip from stacked profiles, but that changes the one-piece construction requirement. Neither process is automatically faster without estimating the actual jobs.
From Choosing a Fabrication Process
Separate slot clearance from kerf
A measured sheet edge is 3.02 mm thick. Its receiving slot needs 0.10 mm total clearance across the width. Assume a constant 0.20 mm kerf centered on the path and no automatic compensation. What path-to-path slot width produces the desired finished slot? Give millimeters to the nearest 0.01 mm. These are exercise values, not a recommended fit.
Show worked solution
First choose the finished opening: 3.02 + 0.10 = 3.12 mm. Then compensate the opening's kerf: 3.12 − 0.20 = 2.92 mm. The cut enlarges this path width to 3.12 mm. Clearance is added once because the question gives total clearance, not clearance per side.
See Laser Cutting and Engraving from CAD for the path-offset method.
From Dimension Anatomy and Placement
Identify the extension line. Which description identifies an extension line?
- The line with arrows at both ends showing a measured span.
- The thin line projecting a feature endpoint toward a dimension.
- The heavier line outlining the part.
Show the reasoning
Choice 2. The extension line carries a feature endpoint out to the dimension line. The dimension line carries the arrows that show the measurement span.
From Dimension Anatomy and Placement
Place the short dimension. A tab has a .75 width dimension and the part has a 3.60 overall width. Both dimensions are above the view. Which order is clearer?
- Put .75 nearer the part and 3.60 farther out.
- Put 3.60 nearer the part and .75 farther out.
- Put both dimension lines on the top edge.
Show the reasoning
Choice 1. Put the short dimension nearest the part so its extension lines need not cross the overall dimension line.
From Dimension Anatomy and Placement
Separate paper spacing from part scale. A drawing is plotted at half size (1:2). Using the 10 mm first-dimension layout guide, how far should the first dimension line be from the outline on the final sheet? Enter a whole number in mm.
Show the reasoning
10 mm. This guide describes paper spacing, so do not halve it to 5 mm or double it to 20 mm.
From Choosing Dimensions
Find the extra center-spacing requirement. Two holes are located from the same left edge: the first at .70 in and the second at 2.55 in. The design requires those two edge-based locations. Which additional controlling dimension repeats information?
- Hole diameter ⌀.28.
- Plate thickness .45.
- Center-to-center distance 1.85.
Show the reasoning
Choice 3. 2.55 − .70 = 1.85 in. That spacing closes the location chain. Leave it out or show (1.85) for reference if useful. The hole diameter and plate thickness supply different information.
From Choosing Dimensions
Read the parentheses. A spacing is shown as (2.40). What does the pair of parentheses tell the reader?
- The actual spacing must be exactly 2.40 with no variation.
- The value is reference information.
- The value is in millimeters even on an inch drawing.
Show the reasoning
Choice 2. It is reference information. It does not impose another acceptance requirement; the actual spacing is controlled by other requirements.
From Choosing Dimensions
Count the simple plate dimensions. A rectangular solid has one off-center circular through hole normal to its broad face. No special symmetry or other features are given. Count the independent numeric dimensions for nominal geometry: body sizes, hole size and hole location. Do not count THRU or tolerance specifications.
Show the reasoning
6 dimensions: body length, height and thickness (3); hole diameter (1); and two center coordinates (2). THRU states the end condition. This count is for the stated ideal geometry, not a universal count for every production drawing.
From Choosing Dimensions
Count dimensions for two equal holes. A rectangular solid has two equal-diameter through holes at unrelated positions, normal to its broad face. No symmetry or pattern spacing is given. Count body dimensions, the shared hole-size value and the four center coordinates. Do not count 2X, THRU or tolerances.
Show the reasoning
8 dimensions: 3 for the body + 1 shared diameter + 2 coordinates for each of 2 centers = 8. A 2X note shares the size; it does not share the center location.
From Choosing Dimensions
Choose the view for the hole diameter. A plate’s front view shows the round hole as a visible circle; its edge view shows the hole as two dashed hidden lines. Where should its diameter callout go?
- On the visible circle in the front view.
- Between the hidden lines in the edge view.
- On both views as separate controlling dimensions.
Show the reasoning
Choice 1. The visible circular outline gives the reader the clearest connection between the leader and the round feature.
From Hole and Feature Callouts
Separate the main hole from the recess. An inch drawing gives 3X ⌀.312 THRU. What is the main hole diameter? Give three decimal places.
⌴ ⌀.625 ↧ .187
Show the reasoning
.312 in. The first diameter belongs to the main through hole. .625 is the counterbore diameter; .187 is its depth. 3X counts the repeated holes.
From Hole and Feature Callouts
Read the counterbore diameter. An inch drawing gives 3X ⌀.312 THRU. What is the counterbore diameter? Give three decimal places.
⌴ ⌀.625 ↧ .187
Show the reasoning
.625 in. Read the diameter immediately following the counterbore symbol. Do not use the .312 main-hole diameter.
From Hole and Feature Callouts
Read the counterbore depth. An inch drawing gives 3X ⌀.312 THRU. How far below the entry surface is the counterbore floor? Give three decimal places.
⌴ ⌀.625 ↧ .187
Show the reasoning
.187 in. The depth symbol follows the counterbore diameter. The main hole continues through; the recess stops at this depth.
From Hole and Feature Callouts
Find the full-diameter blind depth. A hole note reads ⌀.24 ↧ .48 on an inch drawing. A pointed drill bottom is shown. What is the stated depth of the full-diameter portion? Give two decimal places.
Show the reasoning
.48 in. The depth is to the end of the cylindrical, full-diameter portion. Do not add an assumed drill-point length to the stated value.
From Hole and Feature Callouts
Read the included angle. Read ⌀.22 THRU / ⌵ ⌀.46 × 82°. What is the included countersink angle? Enter whole degrees.
Show the reasoning
82°. This is the angle across both sides of the cone. 41° would be the half-angle, not the specified included angle.
From Hole and Feature Callouts
Read the number of places. A note reads 6X ⌀.18 THRU and points to the identified hole pattern. How many holes share the stated diameter and end condition?
Show the reasoning
6 holes. 6X is a count, not a multiplier applied to the .18 diameter. The drawing must also locate the pattern.
From Hole and Feature Callouts
Calculate slot center spacing. An original straight slot has semicircular ends, .50 in width and 1.90 in overall length. What is the distance between the end centers? Give two decimal places.
Show the reasoning
1.40 in. Subtract the two end radii, or one full width: 1.90 − .50 = 1.40. Subtracting just one radius gives 1.65, which is incorrect. This is a geometry check, not a reason to add a third controlling dimension to the drawing.
From Hole and Feature Callouts
Read a decimal-inch value. The drawing states that dimensions are in inches. A feature is labeled .625. Enter its value in inches to three decimal places.
Show the reasoning
0.625 in (also written .625). The omitted leading zero follows decimal-inch drawing convention. There is no conversion to millimeters in this question.
From Hole and Feature Callouts
Keep the decimal places. The same nominal length appears as .40 in one drawing and .400 in another. What must you check before treating their requirements as equivalent?
- Whether .400 is ten times larger.
- Each drawing’s applicable tolerance block and local tolerance.
- Whether extra zeros make the actual part mathematically exact.
Show the reasoning
Choice 2. Their nominal lengths are equal, but the general tolerance blocks or local tolerances may assign different limits. No universal tolerance can be inferred from the number of decimal places alone.
From Engineering Graphics in the Design Process
A graphic for brainstorming: A team wants to compare three possible handle arrangements before choosing dimensions. Which graphic best supports a quick discussion?
- A fully released drawing set
- Labeled ideation sketches
- A completed inspection report
Show answer and reasoning
Choice 2. Quick sketches let the team compare alternatives without pretending that dimensions and requirements are final.
From Engineering Graphics in the Design Process
A graphic for a fit review: You need to examine how several proposed components occupy the available space. Which is the most useful starting graphic?
- An exploded view of the assembly
- A detail drawing of the largest part
- A CAD assembly model
Show answer and reasoning
Choice 3. An assembly model represents component geometry and arrangement. It supports a fit review, but does not by itself prove every functional requirement. An exploded view separates the components on purpose, so it cannot show whether they clash in their assembled positions.
From Engineering Graphics in the Design Process
The governing convention: A drawing arrives from another country. How should you decide its projection convention?
- Read its stated projection method and governing requirements
- Assume every drawing from that country is identical
- Use whichever arrangement is familiar
Show answer and reasoning
Choice 1. Country is not a substitute for the drawing requirements. Resolve a missing or conflicting declaration before using the drawing.
From Working Drawings and Title Blocks
A drawing for manufacture: A supplier needs to make and inspect one custom spacer. Which drawing provides the part requirements?
- An assembly overview with no part dimensions
- An unannotated pictorial image
- A complete detail (part) drawing
Show answer and reasoning
Choice 3. The part drawing defines the spacer through the necessary geometry, dimensions, tolerances, material, and notes.
From Working Drawings and Title Blocks
A drawing for assembly: A technician needs to identify components and see how they fit together. Which drawing should lead the package?
- An assembly drawing with the required component identification
- A material certificate alone
- Only an isolated part drawing
Show answer and reasoning
Choice 1. The assembly drawing communicates relationships among components. Supporting part drawings or instructions may also be required.
From Working Drawings and Title Blocks
Find the dimension units: A drawing shows a length value of 40 without a unit beside it. Where should you first look for the drawing-wide unit declaration?
- The revision description
- The units field or general units note
- The sheet count
Show answer and reasoning
Choice 2. The units declaration explains how unlabeled dimension values are interpreted. Check for any explicitly stated exceptions.
From Working Drawings and Title Blocks
Interpret sheet scale: A view uses the stated sheet scale 1:2. Its dimension reads 60 mm. What is the required nominal length?
- 30 mm
- 120 mm
- 60 mm
Show answer and reasoning
Choice 3. Scale changes the displayed size of the view, not the dimensioned part size. Use the dimension, not a ruler measurement of the image.
From Working Drawings and Title Blocks
Locate the material requirement: Two parts look identical in shaded CAD images. Where should you look to determine the specified material?
- The material field and any referenced material specification
- The shading color alone
- The drawing scale
Show answer and reasoning
Choice 1. Appearance does not establish material. Read the material requirement and its referenced specification.
From Working Drawings and Title Blocks
Confirm the authorized revision: You find two PDFs with the same drawing number but different revisions. Which action determines the correct one for the job?
- Choose the file with the newest download date
- Match the revision to the authorized release record
- Choose the shorter file
Show answer and reasoning
Choice 2. The authorized release record identifies the revision required for the job. A download date is not approval.
From Creating Drawings from CAD Models
Derive a related view: A base view is already linked to the intended model. You need an aligned adjacent view. What is the sound next step?
- Insert a second base view from the model and line it up by eye
- Stretch a copy of the base view
- Create a projected view from the base
Show answer and reasoning
Choice 3. Deriving the projected view preserves the model relationship and intended alignment. A second independent base view shows correct geometry but is not tied to its neighbour, so its alignment and scale can drift when the sheet changes. Check its placement against the projection method.
From Creating Drawings from CAD Models
Change geometry deliberately: A part must be longer. You are considering typing a larger number over a drawing dimension label. What should you do?
- Change the controlling model geometry, then update and inspect the drawing
- Override the label and assume the part changed
- Change only the sheet scale
Show answer and reasoning
Choice 1. A text override can misrepresent unchanged geometry. Use the model control and inspect the linked drawing dimensions after updating.
From Creating Drawings from CAD Models
Review a model change: A referenced model changes after you exported a drawing PDF. What should happen before issuing the revised drawing?
- Send the old PDF because all exports update automatically
- Update the drawing as needed, review affected dimensions and notes, then export and check the approved revision
- Change only the PDF filename
Show answer and reasoning
Choice 2. Associativity helps update the native drawing, subject to its references and settings. It does not review the drawing or update an already exported PDF.
Estimate an inch cutting-tap drill. Estimate the cutting-tap drill for a 5/16-18 thread using major diameter minus 1/TPI. Give the mathematical estimate to four decimal places in inches, before choosing a stock drill.
Show explanation
5/16 = 0.3125 in; pitch = 1/18 in. Subtracting gives 0.256944… in, or 0.2569 in. The number 18 is TPI, not a length. Starting with a minor diameter would subtract thread depth twice.
Estimate a metric cutting-tap drill. Estimate the cutting-tap drill for M10 × 1.5 using major diameter minus pitch. Give the mathematical estimate to two decimal places in millimeters.
Show explanation
The nominal major diameter is 10 mm and pitch is 1.5 mm. 10 − 1.5 = 8.50 mm. Taking 1/1.5 treats a pitch as though it were TPI and gives the wrong dimension.
Reason about thread percentage. Two holes will receive the same cutting tap. Within the permitted hole-size range, what generally happens when the prepared hole diameter increases?
Show explanation
Thread height decreases and tapping torque generally decreases. The larger hole leaves less material for the tap to cut. Engagement length is an axial distance and is a separate design choice; thread percentage is not a strength percentage.
Choose the correct tap-drill reference. A process changes from a cutting tap to a forming tap for the same thread. Which preparation is appropriate?
Show explanation
Use the forming-tap guidance. Forming displaces material rather than removing chips and generally needs a larger starting hole. The cutting-tap approximation is not a forming-tap specification.
From Thread Callouts and Threaded Fasteners
Read an internal Unified thread. Which reading of 3/8-16 UNC-2B is correct?
Show explanation
0.375 in nominal major diameter, 16 TPI, Unified coarse, class 2 internal. B identifies an internal thread. The major diameter is not the tap-drill diameter, and thread class is not strength grade.
From Thread Callouts and Threaded Fasteners
Read an external fine thread. Which reading of 3/8-24 UNF-3A is correct?
Show explanation
0.375 in nominal major diameter, 24 TPI, Unified fine, class 3 external. A identifies the external thread. The class concerns dimensional fit; it is not the fastener material grade.
From Thread Callouts and Threaded Fasteners
Read a metric thread pitch. What does M8 × 1.25 tell you?
Show explanation
Metric thread, 8 mm nominal major diameter and 1.25 mm pitch. The callout shown does not state a tolerance class. The pitch is a spacing, not a count per inch.
From Thread Callouts and Threaded Fasteners
Convert TPI to pitch. A single-start thread has 20 threads per inch. What is its axial pitch? Give inches to three decimal places.
Show explanation
Pitch = 1/20 = 0.050 in. Because this example is single-start, its lead is also 0.050 in per revolution. TPI itself is not a travel distance.
From Drilling on a Drill Press or Mill
Calculate drill RPM from an inch diameter. For a calculation exercise, assume a cutting speed of 60 ft/min and a drill diameter of 0.375 in. Use the exact-π RPM relationship in Feeds and Speeds, not the rounded shortcut. Give the theoretical speed to the nearest whole RPM. The assumed cutting speed is not a recommendation for an unidentified material.
Show explanation
Using the surface-speed relationship, substitute 60 ft/min and 0.375 in with the inch-to-foot conversion. The result is 611.15498… rpm, or 611 rpm. Use the cutting diameter, not a thread minor diameter or the radius. A real setup still needs suitable cutting data and an available spindle setting.
From Drilling on a Drill Press or Mill
Calculate drill RPM from a metric diameter. For a calculation exercise, assume 24 m/min surface speed and an 8 mm drill. Use the exact-π relationship in Feeds and Speeds. Give theoretical RPM to the nearest whole number. These inputs are assumed, not a material recommendation.
Show explanation
Using the surface-speed relationship with consistent length units gives 954.92966… rpm, or 955 rpm. Forgetting the meter-to-millimeter conversion produces an answer a thousand times too small.
From Reaming, Boring and Hole Seats
Recess a cylindrical screw head. The drawing calls for a cylindrical recess with a flat shoulder around a smaller through hole, so a screw head can sit below the surface. Which operation matches?
Show explanation
Counterbore. Its cylindrical recess and flat shoulder provide the required head space. A countersink would create a cone.
From Reaming, Boring and Hole Seats
Finish a nearly sized pin hole. A hole is on location and has the recommended small finishing allowance. The goal is its final diameter and improved surface finish for a pin. Which operation best matches this plan?
Show explanation
Ream. A reamer removes a small allowance to improve size and finish. It should not be expected to relocate a badly misplaced hole.
From Reaming, Boring and Hole Seats
Make a local bearing seat. A rough surface around a hole needs a shallow flat area for a washer to bear on. No deep head recess is required. Which operation best describes the purpose?
Show explanation
Spot face. It creates a local flat bearing surface. A counterbore may use similar tooling but serves a specified recess.
From Reaming, Boring and Hole Seats
Correct an existing hole axis. An existing hole needs its axis corrected relative to the machine setup. Enough material remains for the intended corrected bore to clean up. Which operation offers control of axis and diameter?
Show explanation
Bore. The boring tool path can be set from the machine axis. A reamer tends to follow the existing hole.
From Reaming, Boring and Hole Seats
Make a conical screw seat. The specified flat-head screw needs a matching conical seat. Which operation makes that feature?
Show explanation
Countersink. Select its angle and diameter from the specified head and drawing; the thread size alone does not specify the seat.
From Electric Motor Types
Hold without feedback: An energized mechanism must hold at commanded steps without a feedback sensor, within its load rating. Which is the best fit? 1 = ordinary brushed DC motor alone; 2 = stepper with driver; 3 = servo system. Enter the choice number.
Show worked solution
2: stepper with driver. It can hold a commanded step while energized. Excessive load can cause missed steps, and open-loop control cannot verify position.
From Electric Motor Types
Electronic commutation: A fan motor has permanent magnets on its rotor, electronic phase switching, and no brushes to wear. Identify it: 1 = brushed DC; 2 = squirrel-cage induction; 3 = brushless permanent-magnet motor. Enter the choice number.
Show worked solution
3: brushless permanent-magnet motor. The electronic drive switches stator current. No brushes alone would not uniquely identify it: induction motors and steppers can also be brushless.
From Electric Motor Types
Simple DC operation: A low-cost motor must rotate from a DC supply without an external electronic commutation drive. Periodic brush wear is acceptable. Choose: 1 = permanent-magnet brushed DC; 2 = BLDC without a drive; 3 = open-loop stepper without a drive. Enter the choice number.
Show worked solution
1: permanent-magnet brushed DC. Its commutator and brushes switch rotor current mechanically.
From Electric Motor Types
Correct position error: A machine measures shaft position and adjusts motor drive to reduce the difference from the command. Identify the control: 1 = open-loop stepper; 2 = servo; 3 = fixed DC supply alone. Enter the choice number.
Show worked solution
2: servo. Feedback closes the loop. Servo describes the controlled system, not a single motor construction.
From DC Motor Torque, Speed, and Power
Speed at constant load: Assume a fixed-voltage linear torque-speed curve with stall torque 0.8 N·m and no-load speed 3,000 rpm. What steady speed results at a constant resisting load of 0.2 N·m? Give whole rpm.
Show worked solution
n = 3,000(1 − 0.2/0.8) = 2,250 rpm. One-quarter stall torque leaves three-quarters no-load speed.
From DC Motor Torque, Speed, and Power
Power at the load point: Assume a fixed-voltage linear torque-speed curve with stall torque 0.8 N·m and no-load speed 3,000 rpm. Find mechanical output power at a constant 0.2 N·m load. Give watts to two decimal places; carry intermediate values to at least four significant figures.
Show worked solution
n = 3,000(1 − 0.2/0.8) = 2,250 rpm. ω = 2,250 × 2π/60 = 235.619… rad/s. P = 0.2 × 235.619… = 47.12 W. Using 235.6 rad/s also gives 47.12 W. Multiplying torque by rpm directly gives 450, which is not watts.
From DC Motor Torque, Speed, and Power
Speed of peak power: An ideal linear motor curve has 1.2 N·m stall torque and 3,600 rpm no-load speed. At what speed is shaft output power greatest? Give whole rpm.
Show worked solution
Peak power occurs at half no-load speed: 3,600/2 = 1,800 rpm. At 3,600 rpm the external torque is zero, so shaft output power is zero.
From DC Motor Torque, Speed, and Power
Maximum shaft power: An ideal linear motor curve has 1.2 N·m stall torque and 3,600 rpm no-load speed. What is its maximum mechanical output power? Give watts to one decimal place; carry intermediate values to at least four significant figures.
Show worked solution
At half speed and half torque: n = 1,800 rpm and T = 0.6 N·m. P = 0.6 × (1,800 × 2π/60) = 113.097… = 113.1 W. Stall torque × no-load angular speed would give 452.4 W: those endpoints cannot occur together.
Find the terminal connection. An assembler needs to connect a wire to the correct numbered terminal. Which drawing is the best starting point?
Show explanation
Wiring drawings identify physical terminal connections.
Explain the circuit function. A technician wants to follow component connections to understand electrical operation. Which drawing is the best starting point?
Show explanation
A schematic shows electrical relationships using component symbols.
Recognize the parts. A beginner wants to match visible parts with their appearances in a drawing. Choose a view.
Show explanation
A pictorial emphasizes recognizable appearances.
Explain system functions. A designer wants to explain source, control, current limiting, and light output without drawing each terminal. Choose a view.
Show explanation
A block diagram groups functions and shows their relationships.
Route board copper. A board designer needs to place footprints and route copper traces. Choose a drawing.
Show explanation
The PCB layout defines physical board placement and routing.
Trace a control rung. A controls technician needs to follow conditions arranged in rungs between rails. Choose a drawing.
Show explanation
Ladder diagrams arrange control logic in rungs.
Trace Boolean operations. A designer wants to follow AND and OR operations between logical inputs and outputs. Choose a drawing.
Show explanation
A logic diagram emphasizes gates and logical relationships.
Locate room fixtures. An installer wants to locate fixtures within a room and decode their marks using a legend. Choose a drawing.
Show explanation
An electrical floor plan locates building devices and uses its legend to explain marks.
Follow distribution between floors. A designer wants to see how distribution equipment relates across building levels. Choose a drawing.
Show explanation
A riser emphasizes distribution between levels.
From Reading Electrical Connections
Read an undotted crossing. In the right-hand example below, the legend says only dotted crossings join. Are the horizontal and vertical wires connected at the crossing?
Show explanation
No. Under the stated convention, the undotted crossing keeps the conductors separate.
From Reading Electrical Connections
Read a dotted crossing. In the left-hand example below, the legend says a filled dot joins crossing conductors. Are the horizontal and vertical wires connected?
Show explanation
Yes. The filled dot explicitly joins them.
From Reading Electrical Connections
Count nodes in a wire drawing. Count all nodes in the two separate wire drawings below, taken together. Only dotted crossings join, and there are no off-page connections.
Show explanation
There are 3 nodes: one in the dotted cross, plus two in the undotted cross. Bends and dots do not add nodes.
From Reading Electrical Connections
Identify symbol 1. Which component or connection does this common form represent?
Show explanation
The repeated long and short plates represent cells in a battery. The long plate is positive.
From Reading Electrical Connections
Identify symbol 2. Which component or connection does this common form represent?
Show explanation
The zigzag is a widely taught US resistor form.
From Reading Electrical Connections
Identify symbol 3. Which component or connection does this common form represent?
Show explanation
The rectangle is a widely taught international resistor form.
From Reading Electrical Connections
Identify symbol 4. Which component or connection does this common form represent?
Show explanation
The blade does not touch the second contact, so this switch is open.
From Reading Electrical Connections
Identify symbol 5. Which component or connection does this common form represent?
Show explanation
The circle with a cross is a common lamp symbol.
From Reading Electrical Connections
Identify symbol 6. Which component or connection does this common form represent?
Show explanation
The outward arrows distinguish a light-emitting diode from an ordinary diode.
From Reading Electrical Connections
Identify symbol 7. Which component or connection does this common form represent?
Show explanation
This common sounder form uses curved sound-wave marks. Check the drawing legend for the exact device.
From Reading Electrical Connections
Identify symbol 8. Which component or connection does this common form represent?
Show explanation
The descending bars mark a ground or reference connection. The drawing must establish its meaning; do not assume protective earth.
From Reading Electrical Connections
Identify symbol 9. Which component or connection does this common form represent?
Show explanation
This common fuse form has a line through the rectangular body. Use the legend to distinguish variants.
From Reading Electrical Connections
Identify symbol 10. Which component or connection does this common form represent?
Show explanation
The M identifies a motor in this common circular form.
From Reading Electrical Connections
Identify symbol 11. Which component or connection does this common form represent?
Show explanation
Two equal parallel plates represent an unpolarized capacitor.
From Reading Electrical Connections
Identify symbol 12. Which component or connection does this common form represent?
Show explanation
The bar identifies the cathode. Unlike the LED, this symbol has no light-emission arrows.
From Reading Electrical Connections
Count nodes with S1 open. Count the nodes with S1 open as drawn. Wires are ideal, and there are no off-page connections.
Show explanation
Open S1 gives four conductor groups: B1+–S1, S1–R1, R1–D1 anode, and D1 cathode–B1−. The answer is 4; component bodies and wire bends do not add nodes.
From Reading Electrical Connections
Count nodes with S1 closed. Count the nodes after replacing the open S1 gap with an ideal closed connection. Wires are ideal, and there are no off-page connections.
Show explanation
Open S1 gives four conductor groups: B1+–S1, S1–R1, R1–D1 anode, and D1 cathode–B1−. Closing S1 merges the first two groups, so the answer is 3.
From Current, Voltage, and Resistance
1. Current in a control circuit: A resistor in a control circuit is 470 Ω. What current flows when 24 V is applied across it? Give the answer in milliamperes.
Show worked solution
- Pick the form with the two quantities you have:
I = V/R = 24 V ÷ 470 Ω = 0.051064 A.
Then move the prefix:
0.051064 A × = 51.06 mA
The current is 51.1 mA. Check the units rather than trusting the arithmetic: an ohm is a volt per ampere, so volts divided by ohms leaves amperes. And mind the case — mA is a milliampere, while MA would be a megaampere, a billion times larger.
From Current, Voltage, and Resistance
2. Voltage drop along a cord: A portable tool draws 12 A through an extension cord. Both of the cord’s conductors are in the current path — out to the tool and back — and together they add 0.32 Ω to the circuit. What is the voltage drop across the cord?
Show worked solution
V = IR = 12 A × 0.32 Ω = 3.84 V.
The drop is 3.84 V, and that is 3.84 V that never reaches the tool: the supply is divided between the cord and the load, and the cord takes its share first.
Read the resistance figure carefully before using it. A cord carries the current out on one conductor and back on the other, so what belongs in this calculation is the resistance of the whole path. A table that gives resistance per unit length gives it for a single conductor; use that figure alone for a cord and the answer comes out about half of what a meter across the ends would show.
From Current, Voltage, and Resistance
3. Resistance from a nameplate: A resistive heating element is rated 1,500 W at 120 V. What resistance does that rating imply?
Show worked solution
- The two quantities given are power and voltage, so use
R = V2/P. R = (120 V)2 ÷ 1,500 W = 14,400 ÷ 1,500 = 9.6 Ω.
The rating implies 9.6 Ω. Two steps get to the same place if you prefer: I = P/V = 12.5 A, then R = V/I = 120 ÷ 12.5 = 9.6 Ω.
This is the resistance at the rated operating point. A metal heating element is not the same resistance cold as it is hot, so a meter reading on an unpowered element will not match the nameplate figure exactly.
From Current, Voltage, and Resistance
4. Heat lost in a cable run: A cable run carries 25 A and has a total resistance of 0.18 Ω. How much power is lost as heat in it?
Show worked solution
P = I2R = (25 A)2 × 0.18 Ω = 625 × 0.18 = 112.5 W.
The run loses 112.5 W as heat, spread along its whole length.
Notice the squared current. Halve the load to 12.5 A and the loss does not halve, it quarters, to 28.1 W. That is why a drop that is harmless on a light load becomes a real problem on a heavy one, and why the length and size of a conductor matter more the more current the load draws.
From Wire and Wire Gauge
Resistance of both wires: Use AWG 18 copper with a 50 ft one-way run and an equal return conductor. Find the loop resistance.
Assume solid copper at 20 °C, DC/resistive behavior and no connection resistance. Use d = 0.127 × 92(36 − n)/39 mm, A = πd²/4, ρ = 1.724 × 10−8 Ω·m, and 1 ft = 0.3048 m exactly. Carry at least four significant figures through intermediate steps and report four significant figures (or an integer gauge).
Show worked solution
Count both wires: 2 × 50 × 0.3048 = 30.48 m. For AWG 18, d = 1.024 mm and A = 0.8230 mm² = 8.230 × 10−7 m² (values shown rounded from the formula). R = ρ(2ℓ)/A = (1.724 × 10⁻⁸)(30.48)/(8.230 × 10⁻⁷) = 0.6385 Ω.
Answer: 0.6385 Ω.
From Wire and Wire Gauge
Voltage lost along a run: Use AWG 18 copper with a 50 ft one-way run and an equal return conductor. At 5 A, find the voltage drop.
Assume solid copper at 20 °C, DC/resistive behavior and no connection resistance. Use d = 0.127 × 92(36 − n)/39 mm, A = πd²/4, ρ = 1.724 × 10−8 Ω·m, and 1 ft = 0.3048 m exactly. Carry at least four significant figures through intermediate steps and report four significant figures (or an integer gauge).
Show worked solution
Count both wires: 2 × 50 × 0.3048 = 30.48 m. For AWG 18, d = 1.024 mm and A = 0.8230 mm² = 8.230 × 10−7 m² (values shown rounded from the formula). V = IR = 5 × 0.6385 = 3.192 V (using unrounded resistance).
Answer: 3.192 V.
From Wire and Wire Gauge
Percentage of source voltage lost: Use AWG 18 copper with a 50 ft one-way run and an equal return conductor. At 5 A with a 24 V source, find the percent voltage drop.
Assume solid copper at 20 °C, DC/resistive behavior and no connection resistance. Use d = 0.127 × 92(36 − n)/39 mm, A = πd²/4, ρ = 1.724 × 10−8 Ω·m, and 1 ft = 0.3048 m exactly. Carry at least four significant figures through intermediate steps and report four significant figures (or an integer gauge).
Show worked solution
Count both wires: 2 × 50 × 0.3048 = 30.48 m. For AWG 18, d = 1.024 mm and A = 0.8230 mm² = 8.230 × 10−7 m² (values shown rounded from the formula). Percent drop = 100 × 3.192/24 = 13.30%. Use source voltage as the denominator.
Answer: 13.30 %.
From Wire and Wire Gauge
Voltage remaining at the load: Use AWG 18 copper with a 50 ft one-way run and an equal return conductor. At 5 A with a 24 V source, find the load voltage.
Assume solid copper at 20 °C, DC/resistive behavior and no connection resistance. Use d = 0.127 × 92(36 − n)/39 mm, A = πd²/4, ρ = 1.724 × 10−8 Ω·m, and 1 ft = 0.3048 m exactly. Carry at least four significant figures through intermediate steps and report four significant figures (or an integer gauge).
Show worked solution
Count both wires: 2 × 50 × 0.3048 = 30.48 m. For AWG 18, d = 1.024 mm and A = 0.8230 mm² = 8.230 × 10−7 m² (values shown rounded from the formula). Vload = 24 − 3.192 = 20.81 V.
Answer: 20.81 V.
From Wire and Wire Gauge
Smallest wire meeting a drop target: For a 24 V source, 5 A and a 50 ft one-way run, choose the smallest conductor from AWG 10, 12, 14, 16 and 18 that keeps voltage drop below 6%. Enter its AWG number.
Assume solid copper at 20 °C, DC/resistive behavior and no connection resistance. Use d = 0.127 × 92(36 − n)/39 mm, A = πd²/4, ρ = 1.724 × 10−8 Ω·m, and 1 ft = 0.3048 m exactly. Carry at least four significant figures through intermediate steps and report four significant figures (or an integer gauge).
Show worked solution
Count both wires: 2 × 50 × 0.3048 = 30.48 m. For AWG 18, d = 1.024 mm and A = 0.8230 mm² = 8.230 × 10−7 m² (values shown rounded from the formula). The drops for AWG 10, 12, 14, 16 and 18 are 2.081%, 3.309%, 5.261%, 8.365% and 13.30%. AWG 14 passes; the next smaller conductor, AWG 16, fails. Smaller conductor means a larger AWG number. The 6% target is chosen for this exercise, not a code limit; ampacity still requires a separate check.
Answer: 14 AWG.
From Battery Capacity and Packs
Energy from capacity: An ideal cell is rated 2.35 Ah at a constant 3.7 V. Find its energy in Wh. Round to four significant figures.
Show worked solution
2.35 Ah × 3.7 V = 8.695 Wh. Ah alone is charge, not energy.
From Battery Capacity and Packs
Watt-hours to joules: Convert 8.695 Wh to joules. Round to four significant figures.
Show worked solution
8.695 × 3600 = 31,302 J, or 31,300 J to four significant figures. A factor of 1000 converts kWh to Wh; it does not convert Wh to J.
From Battery Capacity and Packs
Current from a discharge rating: A 2.35 Ah cell has a maximum continuous discharge rating of 4C. What is its rated current estimate?
Show worked solution
4 h⁻¹ × 2.35 Ah = 9.4 A. This is a rated limit, not a claim that the cell always supplies that current.
From Battery Capacity and Packs
Ideal constant-current runtime: An ideal 4.7 Ah pack supplies a constant 1.8 A. Find runtime in hours, to four significant figures. Assume all rated charge is usable.
Show worked solution
4.7 Ah ÷ 1.8 A = 2.61111… h, or 2.611 h. That is about 156.7 minutes; enter hours.
From Battery Capacity and Packs
Voltage of a 4s3p pack: Matched 3.6 V, 2.5 Ah cells form a 4s3p ideal pack: four cells per series string and three strings in parallel. Find pack voltage.
Show worked solution
4 × 3.6 V = 14.4 V. The three parallel strings do not multiply the voltage.
From Battery Capacity and Packs
Capacity of a 4s3p pack: Matched 3.6 V, 2.5 Ah cells form a 4s3p ideal pack. Find pack capacity in Ah.
Show worked solution
3 × 2.5 Ah = 7.5 Ah. Series adds voltage; parallel adds charge capacity.
From Battery Capacity and Packs
Energy of a 4s3p pack: Matched 3.6 V, 2.5 Ah cells form a 4s3p ideal pack. Find nominal energy in Wh.
Show worked solution
Pack voltage is 14.4 V and capacity is 7.5 Ah. Their product is 108 Wh. Check another way: 12 cells × 3.6 V × 2.5 Ah = 108 Wh.
Choose a primary cell: A clock is designed for a replaceable, non-rechargeable cell. Which category matches that requirement?
Show worked solution
Primary cells are intended for replacement after discharge, not recharging.
Choose a rechargeable battery: A portable instrument must be recharged repeatedly with a compatible charger. Which category is required?
Show worked solution
Secondary batteries are designed for repeated charging and discharging.
Storage using two reservoirs: A stationary storage project already has suitable upper and lower reservoirs. It proposes pumping water uphill and recovering energy downhill. Which method is described?
Show worked solution
Pumped hydro stores gravitational potential energy. This identifies the process; it does not establish a cost advantage at every site.
Storage for a portable instrument: A handheld instrument needs reusable electrical energy storage that travels with it. Which option fits?
Show worked solution
A rechargeable battery fits the portable load. Pumped hydro needs reservoirs and elevation difference.
From Fiber Direction and Elastic Stiffness
Find axial strain: The illustrative lamina has E1 = 40 GPa. A 0° patch carries σx = 10 MPa. What axial strain does the elastic model predict, expressed as a percent?
Show worked solution
Use consistent units: 40 GPa = 40,000 MPa. Then εx = σx/E1 = 10/40,000 = 0.00025. Multiply by 100 to express strain as a percent: 0.025%. Entering 0.00025 would leave the result as a dimensionless ratio rather than the requested percent.
From Fiber Direction and Elastic Stiffness
Find effective axial stiffness: At 45°, the same patch carries 10 MPa axial stress and has axial strain 0.000775. What is Ex = σx/εx in GPa?
Show worked solution
10 MPa / 0.000775 = 12,903.23 MPa = 12.90 GPa. Using 0.0775 as the strain would confuse percent with the dimensionless strain ratio.
From Fiber Direction and Elastic Stiffness
Compare +45° and −45°: For the same unidirectional lamina and uniform axial stress, which statement matches this elastic model? 1 = both axial strain and shear sign reverse; 2 = axial stiffness and axial strain match, while shear strain reverses sign; 3 = +45° is always stronger; 4 = neither orientation develops shear strain.
Show worked solution
2. The axial compliance contains even powers of sine and cosine, so opposite angles have the same axial stiffness. Their extension–shear coupling is mirrored, so engineering shear strain changes sign.
From Fiber Direction and Elastic Stiffness
Interpret the result: The 0° patch stretches least under the same 10 MPa stress. What can this lab conclude? 1 = 0° has the greatest failure strength; 2 = 0° is the safest laminate; 3 = 0° has the greatest modeled axial elastic stiffness for these illustrative constants; 4 = every real composite should use only 0° fibers.
Show worked solution
3. Stress divided by elastic strain gives stiffness. Failure strength, damage, safety and laminate design need evidence and models this one-lamina elastic lab does not contain.
From Fiber Direction and Elastic Stiffness
Convert stress to force: A patch has loaded cross-sectional area 200 mm² and uniform axial stress 10 MPa. Since 1 MPa = 1 N/mm², what total axial force corresponds to that stress?
Show worked solution
F = σA = 10 N/mm² × 200 mm² = 2,000 N. Equal stress means equal force per area; total force changes if loaded area changes.
From Grinding Wheels and Abrasives
Decode all five fields: A supplier defines A 60 J 7 V as aluminum oxide, grit 60, grade J, structure 7 and vitrified bond. Which mapping is correct?
- A = abrasive; 60 = grit; J = grade; 7 = structure; V = bond.
- A = bond; 60 = RPM; J = abrasive; 7 = diameter; V = grade.
- A = abrasive; 60 = grade; J = grit; 7 = bond; V = structure.
Show answer and reasoning
Choice 1. The fields run abrasive, grit, grade, structure, bond. The supplier key defines the codes; no maximum RPM is given here.
From Grinding Wheels and Abrasives
Interpret wheel grade: In that supplier's A 60 J 7 V specification, what does J describe?
- The hardness of the workpiece.
- The resistance to releasing abrasive grains.
- The diameter of each grain.
Show answer and reasoning
Choice 2. Grade describes grain retention by the wheel, not the abrasive mineral's hardness or the workpiece's hardness.
From Grinding Wheels and Abrasives
Interpret wheel structure: A supplier labels the fourth field of A 60 J 7 V as structure 7. What property does this field describe?
- Maximum operating RPM.
- Grain diameter in millimeters.
- Relative spacing between grains.
Show answer and reasoning
Choice 3. Structure describes grain spacing. Interpret its value with the supplier's key.
From Grinding Wheels and Abrasives
Interpret the bond field: A supplier defines V as vitrified in the last field of A 60 J 7 V. What does V identify?
- A glasslike binder.
- A coarse grit size.
- A permissible side load.
Show answer and reasoning
Choice 1. The final field identifies the bond. It does not establish permitted grinding surfaces or speed.
From Grinding Wheels and Abrasives
Compare conventional grit numbers: Two otherwise comparable wheels use grit 36 and grit 80 in the same conventional system. Which statement is the better starting point?
- 80 is coarser and always removes more material.
- 36 and 80 are wheel grades.
- 80 is finer and generally favors smaller scratches.
Show answer and reasoning
Choice 3. A larger conventional grit number means finer grains. Actual finish still depends on dressing, feeds and material.
From Grinding Wheels and Abrasives
Choose a conventional abrasive for HSS: For an HSS bit, which conventional abrasive is the usual starting choice between aluminum oxide and silicon carbide?
- Aluminum oxide.
- Silicon carbide because all harder abrasives suit all metals.
- Either, without checking the wheel specification.
Show answer and reasoning
Choice 1. Aluminum oxide is a general starting choice for steels including HSS. The complete wheel must still suit the operation.
From Grinding Wheels and Abrasives
Choose a conventional abrasive for cast iron: Which is a common conventional abrasive starting choice for gray cast iron?
- Diamond is compulsory for every cast iron.
- Silicon carbide, with a suitable wheel specification.
- No abrasive can cut cast iron.
Show answer and reasoning
Choice 2. Silicon carbide is commonly used for cast iron. This is a family choice, not a complete wheel recommendation.
From Grinding Wheels and Abrasives
Choose a superabrasive for hardened steel: Which superabrasive is the general starting choice for hardened steel?
- Diamond because hardness alone decides compatibility.
- CBN.
- Neither can grind hardened material.
Show answer and reasoning
Choice 2. CBN is commonly used for hardened steels. Diamond is generally associated with carbide and other hard nonferrous materials.
From Grinding Wheels and Abrasives
Choose a superabrasive for cemented carbide: Which superabrasive is a general starting choice for cemented-carbide tool grinding?
- Any steel bench wheel.
- CBN solely because the tool is hard.
- Diamond in a suitable wheel and process.
Show answer and reasoning
Choice 3. Diamond is commonly used for cemented carbide. Wheel bond, shape and grinding conditions still need selection.
From Offhand Grinding
Reject an exceeded RPM limit: A machine runs at 3,450 RPM. A proposed wheel is marked maximum 3,200 RPM. What does the speed comparison establish?
- It passes because the difference is small.
- It fails: machine speed exceeds the printed maximum.
- It passes if the wheel has worn smaller.
Show answer and reasoning
Choice 2. 3,450 > 3,200 by 250 RPM, so this speed combination fails. A smaller diameter does not authorize exceeding the marked RPM limit.
From Offhand Grinding
Separate a speed check from full approval: A machine runs at 1,750 RPM and a wheel is marked maximum 3,000 RPM. What follows?
- The speed comparison passes; other compatibility checks remain.
- Every mounting and application is now approved.
- The machine must run at 3,000 RPM.
Show answer and reasoning
Choice 1. 1,750 is below 3,000. A maximum is a ceiling, not a required operating speed or complete approval.
From Offhand Grinding
Calculate wheel rim speed: An illustrative 8 in diameter wheel turns at 1,800 RPM. Calculate SFPM using π × D × RPM / 12 and round to the nearest whole SFPM. This is a calculation, not a recommended machine setting.
Show worked solution
π × 8 × 1,800 / 12 = 3,769.911… SFPM, or 3,770 SFPM. Omitting the division by 12 gives inches per minute, not feet per minute.
From Offhand Grinding
Calculate rim speed after diameter loss: An illustrative wheel now measures 6.5 in diameter and turns at 1,725 RPM. Calculate its rim speed to the nearest whole SFPM. Do not change the wheel's printed maximum RPM.
Show worked solution
π × 6.5 × 1,725 / 12 = 2,935.426… SFPM, or 2,935 SFPM. Use the current diameter; using radius would halve the result.
From Precision Grinding
Count full-face downfeed increments: An idealized plan removes 0.018 in from one face at 0.0015 in per full-face layer. How many downfeed increments are needed? Assume exact removal, no dressing allowance and no spark-out. One layer can require several table strokes; these are arithmetic values, not settings.
Show worked solution
0.018 / 0.0015 = 12 increments. The units of length cancel. This is not a count of individual table strokes.
From Precision Grinding
Plan a smaller final increment: An idealized plan must remove 0.017 in from one face. Eight full-face layers have each removed exactly 0.002 in. What thickness remains for the final layer? Give inches to three decimal places. Ignore wheel wear and elastic effects; these are arithmetic values, not settings.
Show worked solution
0.017 − (8 × 0.002) = 0.001 in. A ninth full 0.002 in layer would remove too much. Use the smaller remainder in this idealized plan.
From Milling Vises and Strap Clamps
Left reading after a midpoint zero. Across a 100 mm sweep, an ideal straight jaw rises smoothly from 0.000 mm at the left end to +0.060 mm at the right. Moving the jaw toward the indicator body was demonstrated to increase the reading. You move to 50 mm and zero the indicator there. What will the left-end reading be? Give millimetres to three decimal places.
Show reasoning
The midpoint raw reading is +0.030 mm. Zeroing subtracts that value from every position, so the left becomes 0.000 − 0.030 = −0.030 mm. The right becomes +0.030 mm. Right minus left is still +0.060 mm.
From Milling Vises and Strap Clamps
Sweep span after re-zeroing. The same ideal jaw reads −0.030 mm at the left, 0.000 mm at midspan and +0.030 mm at the right after zeroing at midspan. What is right minus left in millimetres?
Show reasoning
+0.030 − (−0.030) = +0.060 mm. Re-zeroing changes every displayed value by the same constant, so pairwise differences stay unchanged.
From Milling Vises and Strap Clamps
What zeroing changes. Which statement describes zeroing an indicator at one position during a comparison sweep?
- It physically straightens the jaw at that position.
- It calibrates the instrument across its full range.
- It stores a reference reading that is subtracted from subsequent displayed readings.
Show reasoning
Choice 3. Zeroing establishes the displayed reference. It does not move the jaw, prove calibration or set the job's acceptance requirement.
From Shop Inspection and Quality Control
GO plug will not enter
A clean, identified plain GO plug will not enter the specified inspection length of a hole using the approved method. Has this check passed? What should happen next?
Self-check: write your reasoning before revealing. This response is not automatically graded or saved.
Reveal reasoning
No. The GO plug should enter. Hold the part for investigation; an undersize hole is one possible cause, but form, burrs or alignment can also obstruct entry. Do not force the gauge or accept the part because the NO-GO end also stays out.
From Shop Inspection and Quality Control
Both plug ends enter
The GO plug enters, and the NO-GO plug also enters beyond the engagement allowed by the plain-hole gauging procedure. Is the hole acceptable on this check?
Self-check: write your reasoning before revealing. This response is not automatically graded or saved.
Reveal reasoning
No. GO entry is required, but the NO-GO end must not enter beyond the permitted engagement. This is a failed upper-limit check; investigate and contain the affected work.
From Shop Inspection and Quality Control
Which ring should pass?
For a plain shaft checked by an appropriate GO/NO-GO ring pair, which ring should pass over the specified length? What does a NO-GO ring passing mean?
Self-check: write your reasoning before revealing. This response is not automatically graded or saved.
Reveal reasoning
The GO ring should pass; it checks the large-shaft limit. The NO-GO ring should not pass under the specified procedure. If it passes, the shaft fails that lower-limit check. Correct gauge identity, condition and use still matter.
From Shop Inspection and Quality Control
Bilateral tolerance decision
An invented size is 18.00 ±0.03 mm. A reading is 18.025 mm. Treat the numbers as exact and include the limits. Enter 1 if within limits, 0 if outside.
Show worked solution
Limits are 17.97 to 18.03 mm. 18.025 lies inside, so enter 1.
From Shop Inspection and Quality Control
Unilateral tolerance decision
An invented size is 12.000 +0.020/−0.000 mm. A reading is 11.995 mm. Treat the numbers as exact and include the limits. Enter 1 if within limits, 0 if outside.
Show worked solution
Limits are 12.000 to 12.020 mm. 11.995 is below the lower limit, so enter 0. Do not incorrectly apply ±0.020.
From Shop Inspection and Quality Control
Limit-dimension endpoint
An invented drawing gives limits of 31.96 and 32.04 mm. A reading is 32.04 mm. Treat the numbers as exact and include the limits. Enter 1 if within limits, 0 if outside.
Show worked solution
The upper endpoint is included, so enter 1. The exercise treats the given value as exact; real acceptance uses its specified uncertainty decision rule.
From Process Capability and Variation
Mean of five readings
Five invented readings are 19.98, 20.00, 20.00, 20.03 and 20.04 mm. Find the mean to 0.01 mm.
Show worked solution
The sum is 100.05 mm. Divide by five: 100.05 / 5 = 20.01 mm. The middle observation alone is not the definition of mean.
From Process Capability and Variation
Range of five readings
Five invented readings are 19.98, 20.00, 20.00, 20.03 and 20.04 mm. Find the range to 0.01 mm.
Show worked solution
Largest minus smallest: 20.04 − 19.98 = 0.06 mm. Do not halve this difference; half-range is a different quantity.
From Process Capability and Variation
Is the sample centred?
Five invented readings are 19.98, 20.00, 20.00, 20.03 and 20.04 mm. The limits are 19.95 to 20.05 mm. Compare the sample mean with the limit midpoint. Does this prove capability?
Self-check: write your reasoning before revealing. This response is not automatically graded or saved.
Reveal reasoning
The mean is 20.01 mm; the midpoint is (19.95 + 20.05) / 2 = 20.00 mm. This sample is 0.01 mm above the midpoint, not exactly centred. Its readings all pass, but five readings cannot establish long-term capability.
From Jigs, Fixtures and Repeatable Location
Three primary supports
A rigid block stays seated on three non-collinear underside supports in the XY plane. How many independent degrees of freedom are constrained by this primary arrangement? Give an exact count.
Show worked solution
3: Z translation and rotations about X and Y. The block can still slide along X and Y and rotate about Z until the side and end locators establish the remaining constraints.
From Jigs, Fixtures and Repeatable Location
Add two side locators
After the primary supports, two separated side contacts locate the block against the XZ plane. How many additional independent degrees of freedom do these side contacts constrain while seated? Give an exact count.
Show worked solution
2: Y translation and rotation about Z. Five independent constraints are now established in total.
From Jigs, Fixtures and Repeatable Location
Complete the 3-2-1 arrangement
A block is seated on three primary supports, against two secondary contacts and one tertiary end stop, all appropriately arranged. How many independent degrees of freedom are constrained in total? Give an exact count.
Show worked solution
3 + 2 + 1 = 6. The final stop establishes X position. Clamps maintain seating; do not double the count for positive and negative directions.
From Jigs, Fixtures and Repeatable Location
Bushing-guided drilling
A device locates and clamps a plate, and a hardened bushing guides the drill into the hole position. Is it a jig or a fixture?
Self-check: write your reasoning before revealing. This response is not automatically graded or saved.
Reveal reasoning
It is a jig because it guides the cutting tool as well as holding and locating the part.
From Jigs, Fixtures and Repeatable Location
Program-guided milling
A plate and clamps locate a casting. The CNC axes guide the milling cutter; the workholding device has no tool-guiding feature. Is the device a jig or a fixture?
Self-check: write your reasoning before revealing. This response is not automatically graded or saved.
Reveal reasoning
It is a fixture. The device locates and holds the work, while the machine controls the cutting tool path.
From Turning Tapers
Diameter change per inch. A taper changes from 1.250 to 0.950 in over an axial length of 4.000 in. Find taper per inch to three decimal places.
Show worked solution
(1.250 − 0.950)/4.000 = 0.075 in/in. Use diameter change, not radius change.
Rounded answer: 0.075 in/in.
From Turning Tapers
Convert taper per inch to per foot. A taper rate is 0.075 in/in. Express it in in/ft to three decimal places.
Show worked solution
12 × 0.075 = 0.900 in/ft. A foot contains twelve inches.
Rounded answer: 0.900 in/ft.
From Turning Tapers
Convert taper per foot to per inch. A taper is specified as 0.600 in/ft. Find taper per inch to three decimal places.
Show worked solution
0.600/12 = 0.050 in/in. Do not substitute 0.600 directly into a formula that requires in/in.
Rounded answer: 0.050 in/in.
From Turning Tapers
Compound taper half-angle. A taper has a diameter change of 0.080 in per inch of axial length. Find the compound travel angle from the spindle axis, to three decimal places in degrees.
Show worked solution
α = atan(0.080/2) = 2.291 degrees. The included angle is about 4.581 degrees; that is not the travel angle.
Rounded answer: 2.291 degrees.
From Turning Tapers
Offset with unequal total and taper lengths. A shallow external taper changes from 1.200 to 1.000 in over l = 3.000 in. Total effective center-support length L is 9.000 in. Use the small-angle offset estimate and round to three decimal places.
Show worked solution
Taper rate = 0.200/3.000 in/in. Offset ≈ (0.200/3.000) × 9.000/2 = 0.300 in. Using only the taper length gives 0.100 in and is wrong.
Rounded answer: 0.300 in.
From Turning Tapers
Offset from a per-foot specification. A taper is 0.720 in/ft and the effective center-support length is 8.000 in. Find the small-angle tailstock offset estimate, to three decimal places.
Show worked solution
TPI = 0.720/12 = 0.060 in/in. Offset ≈ 0.060 × 8.000/2 = 0.240 in. Treating TPF as TPI would give 2.880 in.
Rounded answer: 0.240 in.
Pitch from threads per inch. A single-start thread has 16 threads per inch. Find its axial pitch to four decimal places.
Show worked solution
P = 1/16 = 0.0625 in. Lead equals pitch for this single-start thread.
Rounded answer: 0.0625 in.
Lead of a two-start thread. A two-start thread has axial pitch P = 0.075 in. Find the required axial travel along one helix per revolution, to three decimal places.
Show worked solution
Lead = 2P = 2 × 0.075 = 0.150 in/rev. Using pitch alone gives half the required travel.
Rounded answer: 0.150 in/rev.
Ideal sharp-V radial depth. For an ideal sharp-V 60-degree teaching profile with 20 threads per inch, calculate total radial depth H = (√3/2)P. Round to five decimal places. This is a geometric model, not a finished standard thread or a per-pass setting.
Show worked solution
P = 1/20 = 0.050 in. H = (√3/2) × 0.050 = 0.04330 in. Diameter reduction would be 2H, not H. Real thread flats and root shape change the depth.
Rounded answer: 0.04330 in.
Ideal compound travel from radial depth. For the ideal sharp-V 60-degree, 20-per-inch teaching profile, H = (√3/2)/20 in. Find total actual compound travel at 29 degrees from radial using c = H/cos(29°). Round to five decimal places. Ignore tip truncation and backlash; this is not a production setting.
Show worked solution
H = 0.04330127 in. c = H/cos(29°) = 0.04951 in. Compound travel is longer than its radial component; do not double H or use H alone.
Rounded answer: 0.04951 in.
Classify a work observation 1. A tote of parts is moved to an unnecessary distant staging area and then returned. Classify the needless movement of the tote.
Show worked solution
Transportation follows material movement. The observation concerns the tote, rather than the worker’s reaching or searching.
Classify a work observation 2. Unused spare blanks beyond current needs occupy a storage rack. Classify the excess stock itself.
Show worked solution
Inventory describes the excess stock. Making it could have been overproduction, but this observation is about what is stored.
Classify a work observation 3. A worker repeatedly searches several drawers for a measuring tool. Classify the searching.
Show worked solution
Motion follows unnecessary movement by a person. A labeled location may help; check the effect before adopting it.
Classify a work observation 4. A job is ready, but cannot start until missing specifications arrive. Classify the idle delay.
Show worked solution
Waiting is idle time before work can proceed. The missing information should be investigated.
Classify a work observation 5. A station makes 40 extra pieces with no order or replenishment need. Classify the act of making extra pieces.
Show worked solution
Overproduction means making too much or too early. It can create inventory afterward.
Classify a work observation 6. A part meets every functional and appearance requirement, but receives another unnecessary polishing pass. Classify that pass.
Show worked solution
Over-processing adds work beyond requirements. Required finishing would be a different case.
Classify a work observation 7. A part has the wrong hole pattern and must be scrapped. Classify the failed part and its replacement work.
Show worked solution
Defects are failures to meet requirements, leading to scrap or rework. Inspection alone is not a defective part.
Classify a work observation 8. A team never considers a trained assembler’s relevant suggestions for improving the work. Classify the unused contribution.
Show worked solution
Skills waste means failing to use or develop people’s abilities and ideas.
From Value Stream Mapping
Elapsed time through a sequence. One item takes 8 minutes in operation A, waits 75 minutes, then takes 12 minutes in operation B. There are no other delays or overlapping steps. Find elapsed lead time in whole minutes.
Show worked solution
Lead time = 8 + 75 + 12 = 95 minutes. Process time is only 8 + 12 = 20 minutes.
From Outsourcing
Include the avoidable setup. An original hypothetical one-month order needs 100 acceptable brackets. Making requires a new $300 setup plus $9 per acceptable part, both avoided by buying. Shared rent of $600 is paid unchanged either way; a $200 study paid last month is irrecoverable. Assume spare capacity, no displaced work or alternative income, and no other route-dependent costs or credits. What is the total relevant cost of making? Give dollars to the nearest cent.
Show worked solution
$300 + $9 × 100 = $1,200.00. The setup is relevant because this choice changes whether it is paid, even though it is fixed for the order. The unchanged rent and irrecoverable past study do not change the choice. $900 omits the setup; $1,800 includes unchanged rent; $2,000 also includes the sunk study.
From Outsourcing
Count buying charges once. An original hypothetical one-month order needs 100 acceptable brackets. Buying has a $10 per-part quote, a separate $0.50 incoming check per part and $100 freight charged once for the order. The quote excludes checks and freight. Shared rent of $600 stays unchanged; a past $200 study is irrecoverable. Assume the route delivers all 100 acceptable parts by the deadline, with no other route-dependent costs or credits. What is the total relevant buying cost? Give dollars to the nearest cent.
Show worked solution
($10 + $0.50) × 100 + $100 = $1,150.00. The $10.50 already includes the incoming check, so adding another $50 would double count it. $1,000 is only the quote; $1,050 omits freight; $11,050 incorrectly charges freight for every part.
From Outsourcing
Separate unchanged and sunk costs. A hypothetical one-month make-or-buy order would require a new $300 setup only if made. Shared rent of $600 will be paid unchanged under either route. A $200 study paid last month cannot be recovered under either route. Which treatment belongs in a relevant-cost comparison? Enter the choice number.
- Exclude the setup because it is fixed; charge rent only to make.
- Include the new setup for make; exclude unchanged rent and the irrecoverable study from both relevant-cost totals.
- Include all three charges only for make because they can be assigned to the order.
Show worked solution
Choice 2. The new setup changes with this choice; being fixed does not remove its relevance. The rent is a future cost that does not change. The study is a past irrecoverable cost. An allocation label alone does not show that a payment can be avoided.
From Outsourcing
Limit the conclusion to its evidence. A hypothetical comparison for the same 100 acceptable brackets delivered within one month gives relevant totals of $1,200 to make and $1,150 to buy. Equivalent quality and timely delivery are assumed for the calculation; spare internal capacity with no displaced work or alternative income is also assumed. Neither route's actual suitability has been verified. Which conclusion is supported? Enter the choice number.
- Place the supplier order now because a lower total proves acceptable quality and delivery.
- Make the parts because the $300 setup is fixed and must therefore be excluded from the stated making total.
- Buying is $50 lower under the assumptions; verify quality, capacity and deadline evidence and revise costs if those assumptions fail before committing.
Show worked solution
Choice 3. $1,200 − $1,150 = $50. A conditional cost result does not verify a supplier. New avoidable setup belongs in the comparison. Displaced work, if discovered, may add an opportunity cost; other changed assumptions may alter either route's total.
Order a supported drilling route
A small wooden tag will be drilled while attached to a larger strip that provides the required holding area. Its final cut preserves the reference edge. Put these operations in order: A — final sanding and inspection; B — cut the tag free; C — establish the reference edge and mark the hole; D — drill and inspect the hole while the strip is held in the qualified setup.
Self-check: write your answer, then compare below. This response is not automatically graded or saved.
Show a sensible order
C → D → B → A. Establish the reference before locating the hole. The problem specifies holding on the larger strip, so drill before cutting free. Finish and inspect after the cuts. This order depends on the stated setup; it is not a universal rule for drilling.
Find the missing acceptance criterion
A row says: “Drill two through holes using a drill press, the specified bit, backing board and clamps. Use the qualified setup and its recorded speed. Make the holes look good.” What inspection information is missing?
Self-check; your response is not automatically graded or saved.
Show what must be added
“Look good” is not an acceptance criterion. Add the drawing revision, hole-diameter and location limits, any required edge condition, and inspection tools or a defined comparison method. Do not invent the limits; take them from the part requirements.
Find the missing equipment and setup
A row says: “Make a 30 ±0.5 mm-wide strip. Measure width with calipers at both ends.” It identifies material and drawing revision elsewhere. Why can another person still not repeat the operation from this row?
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Show the missing fields
The row gives a result and an inspection method, but no equipment or route. Add the operation, equipment, tool, reference edge, support and holding method, and the required process settings with units or a qualified setup reference. Equipment selection depends on the material and available approved process.
From Critical Dimensions for the Shop
Choose a dimension from function
A cover must slide over two fixed locating pins. Its decorative outer border has plenty of clearance. Which features would you investigate first when choosing critical dimensions, and why?
Self-check; your response is not automatically graded or saved.
Show the functional choice
Start with the hole diameters and their locations relative to the pins and each other. Those control whether the cover fits. The border still needs dimensions, but a tighter border tolerance does not fix misaligned holes. Derive limits from the mating geometry instead of simply choosing a small number.
From Bills of Materials
Allocate sheet material cost
A sheet costs $35. A part is allocated 180 × 100 mm of a 1000 × 400 mm sheet of the same material and thickness. This allocation already includes cutting waste. What is the material cost per part? Report dollars to the nearest cent.
Show worked solution
Share = (180 × 100) / (1000 × 400) = 0.045. Cost = 0.045 × $35 = $1.575, or $1.58 to the nearest cent. Using the length fraction alone gives $6.30 and ignores the width fraction.
From Bills of Materials
Extend BOM quantities once
An invented small stand uses two made side pieces at $0.85 in allocated material each and four purchased pads at $0.30 each. What is the material and component total for one stand, to the nearest cent? Exclude labor and other costs.
Show worked solution
2 × $0.85 + 4 × $0.30 = $1.70 + $1.20 = $2.90. Adding only the two unit prices gives $1.15 and misses the quantities.
From Making Silicone Molds from a Printed Pattern
Split a total silicone mass
For this arithmetic exercise, a silicone's stated A:B ratio is 10:1 by weight. You need 330 g total of A and B combined. How much B is needed? Report grams to one decimal place. This ratio is not a recommendation for another product.
Show worked solution
There are 10 + 1 = 11 ratio parts. B = 330 / 11 = 30.0 g; A = 300.0 g. Check: 300 + 30 = 330 g. Dividing the total by 10 gives 33 g, which treats the total as A alone.
From Making Silicone Molds from a Printed Pattern
Add B to a measured mass of A
An arithmetic example specifies A:B = 10:1 by weight. You have measured 275 g of A, excluding the cup. What mass of B is needed? Report grams to one decimal place. Follow the actual product's ratio for real mixing.
Show worked solution
B = 275 / 10 = 27.5 g. Total mixed mass is 302.5 g. Dividing by 11 gives 25 g and incorrectly treats the given A mass as the total.
From Critical Dimensions for the Shop
Find a center from the measured diameter
In an original classroom DL-01 revision A example, treat the readings as exact: a hole's measured diameter is 4.46 mm, and the vertical distance upward from bottom reference edge B to the nearest hole edge is 18.17 mm. What is the center's y coordinate from B? Report mm to two decimal places. Use the readings, not the drawing's scale or nominal 4.5 mm diameter.
Show worked solution
y = 18.17 + 4.46/2 = 18.17 + 2.23 = 20.40 mm. Adding the full diameter gives 22.63 mm; using the nominal diameter gives 20.42 mm. The measured center meets the DL-01 y limits of 19.5–20.5 mm.
From Critical Dimensions for the Shop
Compare hole readings at the limits
For this classroom DL-01 revision A comparison, all values are exact mm readings. Diameters must be 4.3–4.7; left x from reference edge L must be 14.5–15.5; right x from L must be 84.5–85.5; each y from bottom reference edge B must be 19.5–20.5. Limits include their endpoints. The left hole has diameter 4.3 and center (x, y) = (14.5, 20.5); the right has diameter 4.7 and center (85.5, 19.5). Compare each diameter and each coordinate with its own limits. Does the 71.0 mm difference between x coordinates change those results? Consider only these hole readings.
Self-check; your response is not automatically graded or saved.
Show the comparisons
All six comparisons meet their limits. Diameters 4.3 and 4.7 are the allowed endpoints. Left x = 14.5 and y = 20.5 are allowed; right x = 85.5 and y = 19.5 are allowed. Their x-coordinate difference is 85.5 − 14.5 = 71.0 mm. The nominal x-coordinate difference is 70 mm, but the separate x limits permit this difference; no additional spacing limit is stated. These results say nothing about unreported outline, thickness or surface checks.
From Critical Dimensions for the Shop
Check locations despite nominal spacing
In this classroom DL-01 revision A case, treat the reported center coordinates as exact mm values: left (x, y) = (15.7, 20.0), right (85.7, 20.0). Both x coordinates are measured horizontally right from left reference edge L; both y coordinates are measured vertically up from bottom reference edge B. The drawing requires left x = 15 ±0.5, right x = 85 ±0.5, and each y = 20 ±0.5 mm. A note says, “70.0 mm between centers, so locations meet the drawing.” Check all four coordinates and explain whether that conclusion follows. This question concerns location only.
Self-check; your response is not automatically graded or saved.
Show the location check
The location requirements are not met. Left x = 15.7 exceeds its 15.5 mm upper limit; right x = 85.7 exceeds its 85.5 mm upper limit. Both y = 20.0 readings meet 19.5–20.5 mm. The pair is shifted right even though 85.7 − 15.7 = 70.0 mm. Spacing compares the holes with each other; each coordinate must also locate its hole from the named reference edge.
From Manufacturing Cost and Volume
Equal process costs. For a batch of acceptable parts, A costs $180 fixed plus $9/part; B costs $930 fixed plus $4/part. Both meet the specification. At what quantity are total costs equal? Give a whole number.
Show worked solution
180 + 9Q = 930 + 4Q; 5Q = 750; Q = 150 parts. Each total is $1,530. Dividing the fixed-cost difference by a sum of variable costs would not equate the totals.
From Manufacturing Cost and Volume
Average unit cost. A batch has a $350 setup and $7.375 variable cost per acceptable part. All 80 parts are acceptable. Find average unit cost to the nearest cent.
Show worked solution
C = 350 + 7.375 × 80 = $940. Unit cost = 940/80 = $11.75/part. The $7.375 variable charge alone omits setup.
From Material Utilization and Cost
Count blanks with cutting allowance. A 1,000 mm bar supplies 100 mm blanks. Each blank needs one 2 mm separating cut, including the last blank. Assume no end trim or other allowances. What is the maximum whole number of blanks?
Show worked solution
Each blank consumes 102 mm. Nine consume 918 mm; ten need 1,020 mm and will not fit. Answer: 9 blanks. There are 82 mm left over, plus 18 mm lost to nine kerfs. Ignoring kerf incorrectly predicts ten blanks.
From Material Utilization and Cost
Useful length as a percentage. Use a 1,020 mm bar to make nine acceptable 110 mm blanks of the same cross-section. Include all purchased stock in the denominator. What is utilization to one decimal place?
Show worked solution
Useful length = 9 × 110 = 990 mm. Utilization = 990/1,020 × 100 = 97.0588…%, or 97.1%. 1,020 − 9 × 110 = 30 mm of the bar lies outside the finished blanks (kerf plus offcut together).
From Material Utilization and Cost
Material BOM roll-up. One assembly needs three plates at $4.80 each and six fasteners at $0.30 each. Those prices already include material waste. Find the material-only BOM total to the nearest cent.
Show worked solution
3 × $4.80 + 6 × $0.30 = $14.40 + $1.80 = $16.20. Processing and assembly costs are not included.
From Manufacturing Process Categories
Cut a board to length
A saw cuts a long wooden board into two shorter pieces. Which category describes this operation?
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Separating. The cut removes material along a narrow path and divides the board.
From Manufacturing Process Categories
Make a hole in plastic
A drill removes chips from a solid plastic plate to make a mounting hole. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Separating. The intended change is a hole made by material removal. The plastic family does not change that classification.
From Manufacturing Process Categories
Cut a sheet-metal outline
A laser cuts an outline through thin metal sheet. Material leaves the narrow cut. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Separating. Heat is the means of cutting; removing material to produce the outline is the goal. This is not conditioning merely because heat is involved.
From Manufacturing Process Categories
Build a spacer in layers
A printer deposits plastic layer by layer to build a spacer. Use the five-category model taught here.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Forming. This course includes additive manufacturing within forming. Removing supports afterward would be a separate separating operation.
From Manufacturing Process Categories
Make a cast planter
A wet concrete mixture fills a mold that establishes a planter’s shape. Classify the shape-making step, before considering later property development.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Forming. The mold establishes the shape. Later curing develops properties and is conditioning.
From Manufacturing Process Categories
Bend an aluminum strip
A press bends an aluminum strip into an L-shaped bracket without cutting it. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Forming. The operation changes shape through deformation. A change in properties may accompany it, but the stated goal is the bend.
From Manufacturing Process Categories
Bond a shelf edge
An adhesive bonds a wooden edge strip to a shelf panel. Classify the assembly operation as a whole.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Fabricating. Separate parts become one assembly. Curing of the adhesive is a conditioning mechanism within that joining operation.
From Manufacturing Process Categories
Attach a metal handle
Bolts secure a metal handle to a plastic housing through holes that already exist. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Fabricating. The operation joins parts. The presence of holes does not make this step drilling or separating.
From Manufacturing Process Categories
Dry lumber before cutting
Lumber is dried to reduce its moisture content before parts are cut from it. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Conditioning. Moisture content and resulting behavior change. Shrinkage may accompany drying, but making a new outline is not the purpose.
From Manufacturing Process Categories
Relieve stress in glass
A formed glass piece receives a controlled thermal cycle to reduce residual internal stress. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Conditioning. The goal is to change the internal stress condition. A thermal process is not automatically forming.
From Manufacturing Process Categories
Coat a metal housing
Paint is applied to a metal housing for surface protection and color. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Finishing. The desired result is a protective and decorative surface layer.
From Manufacturing Process Categories
Apply a metal surface layer
A thin metal coating is deposited onto an existing part to improve its surface performance. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Finishing. Adding metal does not automatically mean forming; the stated goal is a surface treatment, not building the main shape.
From Manufacturing Process Categories
Sand to a required thickness
A wooden block is too thick. Sanding continues until it reaches the required thickness. Which category is primary, and what other category might seem plausible?
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Separating is primary because the acceptance goal is a dimension reached by material removal. The surface may also become smoother, an effect associated with finishing.
From Manufacturing Process Categories
Remove visible scratches
A correctly sized plastic panel is polished with a fine abrasive to remove visible scratches. Dimensions are not the acceptance goal. Pick the primary category and explain the overlap.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Finishing is primary because surface appearance is the goal. Abrasion also removes some material, so separating describes part of the mechanism.
From Manufacturing Process Categories
Make a curved laminated rail
Thin wooden strips are bonded around a form. The operation is described as making a specified curve. Pick the primary category and name the second category involved.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Forming is primary under the stated shape goal; fabricating describes bonding the layers. If a question focused instead on achieving the bond, fabricating would be a defensible primary category.
From Manufacturing Process Categories
Develop a resin part’s properties
An already shaped resin part is cured to develop its intended material properties. It is not being attached to another part. Pick the primary category.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Conditioning. The chemical state and properties develop through curing. Without a joining goal, fabricating is not the primary category.
From Manufacturing Process Categories
Join frame members with a weld
A weld joins two metal frame members. Heating also changes material near the joint. Pick the primary category and explain why the side effect does not control it.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Fabricating is primary because the intended result is a joint. Local property changes are real, but they are secondary to the assembly goal in this description.
From Manufacturing Process Categories
Harden the surface of a gear
A treatment changes the material near a metal gear’s surface to increase hardness. No decorative or protective coating is deposited. Use the course model.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Conditioning. The main goal is a change in material properties, even though the change is concentrated near the surface. Surface location alone does not make it finishing.
From Material Families
Copper wire core
Classify only the copper core of an insulated wire.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Metals. Copper belongs to the metal family; the surrounding polymer insulation is a separate part with a different classification.
From Material Families
Steel screw
A screw is specified as steel. Which of the four material families contains the screw material?
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Metals. Steel is an iron-based alloy. Its shape or coating does not change the base material’s family.
From Material Families
Transparent acrylic panel
A transparent panel is identified as acrylic, not glass. Pick its material family.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Plastics (polymers). Transparency is a property shared by materials in different families; acrylic is a polymer.
From Material Families
Unreinforced nylon gear
A small gear is made from unreinforced nylon. Classify its material, not the job the gear performs.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Plastics (polymers). A gear can be made from several families; the stated composition determines this answer.
From Material Families
Plywood shelf panel
A shelf panel consists of thin wood layers bonded together. Use the course’s four-family model.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Wood & Composites. Plywood is an engineered wood composite. There is no separate fifth composites family here.
From Material Families
Glass-fiber-reinforced panel
A panel consists of glass fibers held in a cured polymer matrix. Classify the complete reinforced material, not either ingredient alone.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Wood & Composites. This engineered composite belongs in that combined course family even though it contains no wood. The glass fibers alone and the polymer alone would have different family labels.
From Material Families
Porcelain floor tile
A tile is made of fired porcelain. Pick its material family.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Ceramics & Glass. Porcelain is a ceramic. A hard surface does not make it a metal.
From Material Families
Glass jar body
Classify the glass body of a jar, leaving its metal lid out of the question.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Ceramics & Glass. A product can contain several families; this prompt names only the glass body.
From Material Properties
Rain-exposed equipment cover
Choose a candidate family for a light, nonstructural outdoor cover that must resist rain without rusting. Explain why and name one further check. More than one family can be defended.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
One suitable starting point is Plastics (polymers): a suitable outdoor grade can offer low density and will not rust. Check sunlight resistance, temperature, impact and chemical exposure. A corrosion-resistant metal with suitable protection can also be defensible; “metal” alone is not enough. No family label guarantees outdoor life.
From Material Properties
Reduce a handheld enclosure’s mass
A handheld enclosure should be light, moldable and electrically insulating. Choose a candidate family and explain the tradeoff.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Plastics (polymers) is a reasonable starting point because many unfilled grades have low density, mold readily and are electrical insulators. Check impact resistance, temperature and the grade’s actual electrical properties. A thin metal enclosure might be light, but would need a separate strategy for insulation.
From Material Properties
Carry heat away from a warm component
A plate should carry heat away from a warm component. Electrical insulation is not required. Choose a candidate family, say why, and name a next check.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Metals is a reasonable starting point: aluminum or copper can conduct heat well. Check the actual thermal conductivity, operating temperature, dimensions, environment and cost. Some engineering ceramics can also be suitable; the family tendency is a starting point, not an exclusive answer.
From Material Properties
Conduct electric current
A flexible wire core must conduct electric current well and be drawable into a long, thin shape. Choose a family and name two relevant properties.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Metals. High electrical conductivity and ductility support the requirement; copper is a common candidate. The complete wire also needs suitable insulation and a size matched to its service requirements.
From Material Properties
Hard decorative floor surface
A decorative floor surface needs scratch resistance and low sensitivity to routine water exposure. Choose a candidate family and explain one limitation.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Ceramics & Glass is a reasonable starting family, with a suitable ceramic tile as a candidate. Hardness helps resist scratching. Brittleness means impact and support conditions still matter, and water exposure must be considered for the complete tile-and-joint system.
From Material Properties
A stiff panel with limited mass
A large panel must be stiff without excessive mass. A layered or fiber-reinforced construction is acceptable. Choose a candidate family and explain what must be checked.
Self-check: make a choice and explain it, then compare with the reasoning below. Your response is not automatically graded or saved.
Reveal the classification and reasoning
Wood & Composites is a reasonable starting point: plywood or a suitably designed fiber composite can provide useful stiffness for its mass. Check loading direction, layer arrangement, thickness, joints and moisture or temperature exposure. A shaped metal panel can also be defensible; geometry affects part stiffness.
Inch micrometer reading A. A 0–1 in micrometer has 0.025 in sleeve spaces and 0.001 in thimble spaces. The last reached sleeve line is 0.350 in. Thimble line 18 aligns with the datum. What is the reading in inches?
Show worked solution
Sleeve 0.350 + thimble 0.018 = 0.368 in. At thimble zero, do not add a full extra turn.
Inch micrometer reading B. A 0–1 in micrometer has 0.025 in sleeve spaces and 0.001 in thimble spaces. The last reached sleeve line is 0.375 in. Thimble line 24 aligns with the datum. What is the reading in inches?
Show worked solution
Sleeve 0.375 + thimble 0.024 = 0.399 in. At thimble zero, do not add a full extra turn.
Inch micrometer reading C. A 0–1 in micrometer has 0.025 in sleeve spaces and 0.001 in thimble spaces. The last reached sleeve line is 0.400 in. Thimble line 0 aligns with the datum. What is the reading in inches?
Show worked solution
Sleeve 0.400 + thimble 0.000 = 0.400 in. At thimble zero, do not add a full extra turn.
Inch micrometer reading D. A 0–1 in micrometer has 0.025 in sleeve spaces and 0.001 in thimble spaces. The last reached sleeve line is 0.000 in. Thimble line 0 aligns with the datum. What is the reading in inches?
Show worked solution
Sleeve 0.000 + thimble 0.000 = 0.000 in. At thimble zero, do not add a full extra turn.
Inch micrometer reading E. A 0–1 in micrometer has 0.025 in sleeve spaces and 0.001 in thimble spaces. The last reached sleeve line is 0.975 in. Thimble line 24 aligns with the datum. What is the reading in inches?
Show worked solution
Sleeve 0.975 + thimble 0.024 = 0.999 in. At thimble zero, do not add a full extra turn.
Vernier micrometer reading A. On a 0–1 in vernier micrometer, the last reached sleeve line is 0.450 in. The datum is between thimble 12 and 13. Sleeve vernier line 7 aligns with a thimble line. Each sleeve vernier division adds 0.0001 in. Read the instrument.
Show worked solution
0.4500 + 0.0120 + 0.0007 = 0.4627 in. Keep the completed thousandth; then add the vernier fraction.
Vernier micrometer reading B. On a 0–1 in vernier micrometer, the last reached sleeve line is 0.275 in. The datum is between thimble 24 and 0. Sleeve vernier line 9 aligns with a thimble line. Each sleeve vernier division adds 0.0001 in. Read the instrument.
Show worked solution
0.2750 + 0.0240 + 0.0009 = 0.2999 in. Keep the completed thousandth; then add the vernier fraction.
Vernier micrometer reading C. On a 0–1 in vernier micrometer, the last reached sleeve line is 0.300 in. The datum is exactly at thimble 0. Sleeve vernier line 0 aligns with a thimble line. Each sleeve vernier division adds 0.0001 in. Read the instrument.
Show worked solution
0.3000 + 0.0000 + 0.0000 = 0.3000 in. Keep the completed thousandth; then add the vernier fraction.
Vernier micrometer reading D. On a 0–1 in vernier micrometer, the last reached sleeve line is 0.000 in. The datum is between thimble 0 and 1. Sleeve vernier line 1 aligns with a thimble line. Each sleeve vernier division adds 0.0001 in. Read the instrument.
Show worked solution
0.0000 + 0.0000 + 0.0001 = 0.0001 in. Keep the completed thousandth; then add the vernier fraction.
Vernier micrometer reading E. On a 0–1 in vernier micrometer, the last reached sleeve line is 0.975 in. The datum is between thimble 24 and 0. Sleeve vernier line 9 aligns with a thimble line. Each sleeve vernier division adds 0.0001 in. Read the instrument.
Show worked solution
0.9750 + 0.0240 + 0.0009 = 0.9999 in. Keep the completed thousandth; then add the vernier fraction.
Vernier caliper reading A. A 25-division inch vernier caliper has 0.025 in main-scale spaces and a 0.001 in least count. Its zero is just past the 1.325 in main-scale line. Vernier division 17 aligns. What is the reading?
Show worked solution
Main scale 1.325 + 17 × 0.001 = 1.342 in. Read the main scale at the vernier zero, not at the matching line.
Vernier caliper reading B. A 25-division inch vernier caliper has 0.025 in main-scale spaces and a 0.001 in least count. Its zero is just past the 1.325 in main-scale line. Vernier division 24 aligns. What is the reading?
Show worked solution
Main scale 1.325 + 24 × 0.001 = 1.349 in. Read the main scale at the vernier zero, not at the matching line.
Vernier caliper reading C. A 25-division inch vernier caliper has 0.025 in main-scale spaces and a 0.001 in least count. Its zero is exactly on the 1.350 in main-scale line. Vernier division 0 aligns. What is the reading?
Show worked solution
Main scale 1.350 + 0 × 0.001 = 1.350 in. Read the main scale at the vernier zero, not at the matching line.
Vernier caliper reading D. A 25-division inch vernier caliper has 0.025 in main-scale spaces and a 0.001 in least count. Its zero is exactly on the 0.000 in main-scale line. Vernier division 0 aligns. What is the reading?
Show worked solution
Main scale 0.000 + 0 × 0.001 = 0.000 in. Read the main scale at the vernier zero, not at the matching line.
Vernier caliper reading E. A 25-division inch vernier caliper has 0.025 in main-scale spaces and a 0.001 in least count. Its zero is just past the 0.000 in main-scale line. Vernier division 24 aligns. What is the reading?
Show worked solution
Main scale 0.000 + 24 × 0.001 = 0.024 in. Read the main scale at the vernier zero, not at the matching line.
Metric half-millimeter mark. A metric micrometer has 0.5 mm sleeve spaces and 0.01 mm thimble spaces. The 11 mm line and the next half-millimeter mark are exposed. Thimble 36 aligns. Read the tool in mm.
Show worked solution
11 + 0.5 + 0.36 = 11.86 mm. Missing the half-millimeter line would give 11.36 mm.
Beam plus dial. An inch dial caliper is marked 0.100 in per revolution with 100 dial spaces. The completed beam reading is 2.400 in and the dial indicates 63. Read the caliper.
Show worked solution
2.400 + 63 × 0.001 = 2.463 in. The beam already accounts for completed dial turns.
From Choosing Shop Measuring Tools
Tool for a tight outside size. An accessible outside diameter is 0.750 ±0.0005 in. Use the rule of thumb that resolution should be about one tenth of the TOTAL tolerance. Which candidate meets that resolution target? Enter its number: 1 = 0.01 in steel rule; 2 = 0.001 in caliper; 3 = 0.0001 in outside micrometer. Assume the tools have adequate range and are in good condition.
Show worked solution
Total tolerance = 0.001 in; target increment = 0.0001 in. Choose 3, the outside micrometer. Its accuracy and method still need verification; this resolution rule is not an acceptance standard.
From Choosing Shop Measuring Tools
Tool geometry for a depth. A wide flat-bottom recess is 0.500 ±0.001 in deep. The reference rim is wide enough to support a depth-micrometer base. Which candidate meets the one-tenth-of-total-tolerance resolution target AND reaches this feature? Enter 1 = outside micrometer, 0.0001 in; 2 = depth micrometer, 0.0001 in; 3 = caliper depth rod, 0.001 in. Assume adequate range and verified condition.
Show worked solution
Total tolerance = 0.002 in; target = 0.0002 in. Choose 2. The depth micrometer has both suitable geometry and a fine enough increment. The outside micrometer cannot make this depth measurement.
From Choosing Shop Measuring Tools
Resolution target from bilateral limits. A shaft size is 1.250 ±0.002 in. Using the one-tenth-of-TOTAL-tolerance rule of thumb, calculate the target reading increment in inches. This asks for an increment, not an accuracy guarantee.
Show worked solution
Total tolerance = 0.004 in. Divide by 10: 0.0004 in. Dividing only the ± amount by 10 would use half the stated tolerance width.
From Shop Layout and Angle Measurement
Bevel protractor reading A. A 5-minute vernier bevel protractor shows 26 completed degrees read to the right of zero. Division 7 on the same-side vernier aligns. Express the complete scale reading in TOTAL arcminutes (60 arcminutes = 1 degree).
Show worked solution
26° + 7 × 5′ = 26°35′. In total minutes: 26 × 60 + 35 = 1595 arcmin. Read the vernier in the same direction as the main scale.
From Shop Layout and Angle Measurement
Bevel protractor reading B. A 5-minute vernier bevel protractor shows 48 completed degrees read to the left of zero. Division 11 on the same-side vernier aligns. Express the complete scale reading in TOTAL arcminutes (60 arcminutes = 1 degree).
Show worked solution
48° + 11 × 5′ = 48°55′. In total minutes: 48 × 60 + 55 = 2935 arcmin. Read the vernier in the same direction as the main scale.
From Shop Layout and Angle Measurement
Bevel protractor reading C. A 5-minute vernier bevel protractor shows 17 completed degrees read to the left of zero. Division 4 on the same-side vernier aligns. Express the complete scale reading in TOTAL arcminutes (60 arcminutes = 1 degree).
Show worked solution
17° + 4 × 5′ = 17°20′. In total minutes: 17 × 60 + 20 = 1040 arcmin. Read the vernier in the same direction as the main scale.
From Shop Layout and Angle Measurement
Obtuse angle from an acute supplement. The contact arrangement measures the acute supplement of the required obtuse angle. The scale shows 41 completed degrees and vernier division 9 aligns on a 5-minute instrument. Report the REQUIRED obtuse angle in total arcminutes.
Show worked solution
Scale = 41°45′ = 2505 arcmin. Required angle = 180° − 41°45′ = 138°15′. Thus 10800 − 2505 = 8295 arcmin. Subtract only because the problem explicitly identifies the measured supplement.
From Reliable Shop Measurement
Mean of labeled bore readings. Treat these invented classroom local-diameter readings as exact. The bore is 20 mm deep and has stated diameter limits of 19.980 to 20.020 mm. Near, middle, and far are axial stations 2, 10, and 18 mm from the entry face. X and Y are two marked perpendicular diametral directions at each station. The readings are: near/X 20.000, near/Y 20.002, middle/X 20.001, middle/Y 20.003, far/X 20.004 and far/Y 20.026 mm. Calculate the mean of all six readings to 0.001 mm. The mean alone does not decide whether every stated local reading is within the limits.
Show worked solution
The sum is 120.036 mm. Divide by six: 120.036 / 6 = 20.006 mm. That mean is between 19.980 and 20.020 mm, but it does not replace the six labeled readings.
From Reliable Shop Measurement
Largest labeled bore reading. Treat these invented classroom local-diameter readings as exact. The bore is 20 mm deep and has stated diameter limits of 19.980 to 20.020 mm. Near, middle, and far are axial stations 2, 10, and 18 mm from the entry face. X and Y are two marked perpendicular diametral directions at each station. The readings are: near/X 20.000, near/Y 20.002, middle/X 20.001, middle/Y 20.003, far/X 20.004 and far/Y 20.026 mm. Enter the largest stated local reading to 0.001 mm. Keep its location in your notes.
Show worked solution
The largest reading is 20.026 mm at far/Y. It is 0.006 mm above the stated 20.020 mm upper limit. The mean, 20.006 mm, is a different calculation and is not the answer to this local-reading question.
From Reliable Shop Measurement
What the labeled readings support. Treat these invented classroom local-diameter readings as exact. The bore is 20 mm deep and has stated diameter limits of 19.980 to 20.020 mm. Near, middle, and far are axial stations 2, 10, and 18 mm from the entry face; X and Y are two marked perpendicular diametral directions at each station. Six readings have a mean of 20.006 mm, but the far/Y local diameter at 18 mm depth is 20.026 mm. Which conclusion is supported? Enter its number: 1 = the in-limit mean proves the whole bore is within limits; 2 = the sampled far/Y reading is above the stated upper limit, while the unsampled bore and its complete form remain unknown; 3 = the six readings prove a specific taper and roundness error.
Show worked solution
Choose 2. The far/Y sample is above the stated upper limit. Averaging does not erase that result, and six sparse readings do not certify the entire bore or establish a complete taper or roundness condition.
From Forming Metal
A small batch of large housings
A large hollow iron housing is needed in a small batch. Which route is the strongest starting candidate among these?
- Pressure die casting in an aluminum die-casting cell
- Sand casting with a core
- Wire drawing
Show answer and reasoning
Choice 2. Sand casting can accommodate large castings and internal cavities using cores, with tooling suited to varied batch sizes. Confirm alloy, wall design and later machining.
From Forming Metal
Repeated small housings
A stable design needs a long run of detailed zinc housings that can eject from a metal die. Which route deserves investigation?
- Pressure die casting
- Open-die forging
- Wire drawing
Show answer and reasoning
Choice 1. Pressure die casting repeats detailed shapes in suitable alloys. A long run can spread the die cost, but quantity alone does not prove an economic crossover.
From Forming Metal
A long constant profile
A rail has the same intricate cross-section along its length. Local holes will be machined later. Which starting route fits?
- Deep drawing
- Sand casting each short length
- Extrusion
Show answer and reasoning
Choice 3. Extrusion creates a long profile through a shaped die. The later holes do not invalidate that choice; they are separate operations.
From Forming Metal
Twenty bent brackets
Twenty simple sheet brackets are needed, and their design may change. Choose a flexible forming route after the blanks have been cut.
- Dedicated high-volume progressive tooling
- Press-brake bending with suitable tooling
- Wire drawing
Show answer and reasoning
Choice 2. Press-brake bending suits this simple shape and small, changeable batch. Dedicated tooling needs an economic justification; drawing wire makes a different geometry.
From Forming Metal
Classify a stamping operation
A press cuts the required flat outline out of sheet, without bending it. What is this operation?
- Blanking: Separating
- Blanking: Conditioning
- Deep drawing: Forming
Show answer and reasoning
Choice 1. The wanted cut-out is the blank. Cutting separates material even if the machine is part of a stamping line.
From Forming Metal
Casting size versus shrinkage voids
Which feature supplies liquid metal to a casting as it solidifies?
- Draft
- The core
- A properly designed riser
Show answer and reasoning
Choice 3. The riser feeds contraction during solidification. Draft aids release and the core creates an internal opening; neither replaces feeding.
From Joining Metal
Join repeated overlapping sheets
Many overlapping steel-sheet assemblies need discrete permanent joints. Both sides are accessible. Which process fits these conditions?
- Resistance spot welding
- Soldering simply because it uses less heat
- Extrusion
Show answer and reasoning
Choice 1. Resistance spot welding forms local nuggets at overlapping sheets and suits repeated production. Separate spots are not a continuous leak-tight seam.
From Fasteners and Galvanic Corrosion
A removable service cover
A cover must be removed repeatedly for maintenance. Tool access exists. Which starting joint makes most sense?
- A continuous fusion weld
- Compatible threaded fasteners
- Solid rivets chosen for permanent retention
Show answer and reasoning
Choice 2. Threaded fasteners support disassembly. The design still needs suitable engagement, strength, access and environmental compatibility.
From Joining Metal
Join without melting the base parts
A suitable dissimilar-metal joint uses a molten filler drawn into a narrow gap while both base parts remain solid. Which process family matches?
- Fusion welding only
- Rolling
- Brazing or soldering
Show answer and reasoning
Choice 3. Brazing and soldering melt filler rather than the base parts. Filler temperature regime, compatibility and service requirements distinguish the choice.
From Fasteners and Galvanic Corrosion
Spot the galvanic risk
Assume two metals have sufficiently different electrochemical behavior in salt water. Which arrangement completes a galvanic cell?
- They share salt water and electrical contact
- They are dry and electrically isolated
- They are in separate sealed dry boxes
Show answer and reasoning
Choice 1. The first arrangement supplies both electrolyte and electrical connection. Dissimilar names alone do not complete a cell.
From Fasteners and Galvanic Corrosion
A washer does not isolate the whole joint
An insulating washer is under a bolt head, but the bare bolt shank touches the dissimilar plate in a wet hole. What should the designer conclude?
- The washer proves isolation
- An electrical path remains; review complete isolation and drainage
- All corrosion has been prevented
Show answer and reasoning
Choice 2. The shank contact bypasses the washer. A compatible sleeve and washers may be part of a complete design, subject to load and environmental requirements.
From Fasteners and Galvanic Corrosion
Decode an inch thread
What does the thread-size designation 1/4-20 UNC communicate?
- Nominal 1/4-inch thread diameter, 20 threads per inch, Unified coarse series
- 1/4-inch bolt length and a 20-inch pitch
- 6 mm nominal diameter and 1 mm pitch
Show answer and reasoning
Choice 1. The first value is nominal diameter; 20 is a count of threads over one inch, so pitch is 1/20 inch. UNC names the Unified coarse series. Bolt length and the remaining specification are separate.
From Fasteners and Galvanic Corrosion
Decode a metric thread
What does M6 × 1 mean, and does it match 1/4-20 UNC?
- 6 mm wrench size and one thread per inch; it matches
- 6 mm bolt length and 1 mm diameter; it matches
- 6 mm nominal thread diameter and 1 mm pitch; it does not match
Show answer and reasoning
Choice 3. M6 identifies nominal 6 mm thread diameter; 1 is the axial spacing in millimeters. An inch designation uses threads per inch instead. These two sizes differ in diameter and spacing; neither designation specifies wrench size or bolt length.
From Joining Metal
Classify the filler temperature
A close-fitting joint is filled by capillary action. The filler has a liquidus of 600°C and both base parts remain solid. Which process matches?
- Soldering because the base parts stay solid
- Brazing because the filler liquidus is above 450°C
- Fusion welding because some metal becomes liquid
Show answer and reasoning
Choice 2. The filler liquidus, not the flame temperature, is compared with 450°C. At 600°C it is in the brazing regime. Both brazing and soldering keep the base parts solid; fusion welding melts base metal at the joint.
From Fasteners and Galvanic Corrosion
Compare wetted galvanic areas
Compare the same active/noble metal pair in the same electrolyte with electrical contact. Which arrangement can concentrate galvanic attack on the active metal most strongly?
- A large wetted active area coupled to a small wetted noble area
- A dry, fully isolated pair
- A small wetted active area coupled to a large wetted noble area
Show answer and reasoning
Choice 3. The small active anode receives the attack associated with the larger wetted cathode. A coating defect can create that unfavorable area ratio. The dry isolated pair lacks the conditions stated for the working cell; area ratio alone does not predict service life.
From Metals and Their Properties
Choose a property for impact: A part must absorb energy before fracturing during an impact. Which property most directly describes that requirement?
- Density
- Toughness
- Electrical conductivity
Show answer and reasoning
Choice 2: toughness. Hardness and strength alone do not establish resistance to impact fracture.
From Nonferrous Metals
Choose a conductor: Choose a starting material for a compact electrical conductor when conductivity is the main requirement and mass is secondary.
- Copper
- Tool steel
- Gray cast iron
Show answer and reasoning
Choice 1: copper. The final choice still requires a suitable grade and conductor design.
From Types of Steel
Read an incomplete order: An order says only “4140 round bar, 30 mm diameter.” What is still needed to establish the intended mechanical condition?
- The color of the packaging
- A magnet test
- The delivery condition and required properties
Show answer and reasoning
Choice 3. The grade alone does not establish heat treatment or hardness. Length, quantity, tolerances and the governing specification also belong on a complete order.
From Heat Treating Metals
Choose the purpose of tempering: A suitable steel has been quench hardened. What is the usual reason for tempering it?
- Make it molten
- Adjust hardness and toughness after hardening
- Turn it into a nonferrous metal
Show answer and reasoning
Choice 2. Tempering adjusts the properties of hardened steel, commonly sacrificing some hardness to improve toughness.
From Heat Treating Metals
Restore formability: A cold-worked metal needs restored ductility before further forming. Which process is the appropriate starting point for a material-specific treatment plan?
- Annealing
- More cold work
- Painting
Show answer and reasoning
Choice 1. A suitable anneal can restore ductility. More cold work generally uses up remaining ductility; painting changes the surface coating.
From Metal Stock Forms and Purchasing
Calculate solid round bar mass: A hypothetical stock estimate specifies density = 2,700 kg/m³. Use that supplied assumption, not a certified alloy property. A solid round bar has diameter 20 mm and length 500 mm. Find its mass in kg, rounded to three decimal places.
Show answer and reasoning
Radius = 0.010 m and length = 0.500 m. Volume = π × 0.010² × 0.500 = 0.000157080 m³. Mass = volume × 2,700 = 0.424115 kg, or 0.424 kg. Using the diameter as the radius gives 1.69646 kg, four times too much. This is mass, not weight force.
From Metal Stock Forms and Purchasing
Calculate tube material volume: Round tube has outside diameter 30 mm, wall thickness 3 mm and length 200 mm. Find the volume of metal in cm³, rounded to two decimal places. Ignore end features and manufacturing tolerances.
Show answer and reasoning
Inside diameter = 30 − 2 × 3 = 24 mm. Volume = π(30² − 24²) × 200 / 4 = 50,893.801 mm³. Divide by 1,000 to obtain 50.89 cm³. Treating the tube as solid gives 141.37 cm³; subtracting only one wall gives 26.86 cm³.
From Metal Stock Forms and Purchasing
Estimate cost from supplied mass: A supplier quote for an invented example charges $6.50 per kg plus one $4.25 cut charge. The supplied billable mass is 2.90 kg. Find the total before tax and shipping, rounded to cents.
Show answer and reasoning
2.90 × 6.50 + 4.25 = $23.10. Omitting the cut charge gives $18.85. These are supplied practice values, not current market prices.
From Analog-to-Digital Conversion
ADC code from voltage: Use an ideal 10-bit ADC, exact 5.000 V reference and the floor-bin model. What code represents 3.000 V? Round down after multiplying by 1024/5.
Show worked solution
floor(3.000 × 1024 / 5) = floor(614.4) = 614. Using 1023 here would give 613 after rounding down.
From Analog-to-Digital Conversion
Voltage from ADC code: With an exact 5.000 V reference, find the lower-bin-edge voltage for code 800. Use 1024 steps and give four significant figures.
Show worked solution
800 × 5 / 1024 = 3.90625 V, or 3.906 V. Dividing by 1023 gives about 3.910 V, a different endpoint scaling.
From Analog-to-Digital Conversion
ADC step size: For 10 bits and an exact 5.000 V reference, what is the nominal ADC step in mV? Give four significant figures.
Show worked solution
5 / 1024 × 1000 = 4.8828125 mV, or 4.883 mV.
PWM period from frequency: An ideal PWM signal repeats at 490 Hz. Find its period in milliseconds, to four significant figures.
Show worked solution
T = 1/490 s × 1000 = 2.041 ms. Frequency is cycles per second; period is seconds per cycle.
PWM on-time: An ideal PWM signal is 490 Hz with 25% duty. Find its on-time in microseconds. Keep at least four significant figures in intermediate results.
Show worked solution
T = 1/490 = 0.002040816 s. On-time = 0.25 × T × 1000000 = 510.2 µs. Using a rounded T of 2.041 ms gives 510.25 µs.
Average PWM voltage: An ideal output switches between 0 and 5.000 V at 35% duty. What is its average voltage?
Show worked solution
D = 35/100 = 0.35. Average = 0.35 × 5 = 1.750 V. The pin is not steadily at that voltage.
Nominal analogWrite value: Use the nominal mapping round(255 × D). What analogWrite value requests 35% duty on an Uno R3 PWM pin? Round to the nearest integer.
Show worked solution
round(255 × 0.35) = round(89.25) = 89. This is a nominal request, not an exact timer-duty guarantee.
Pin mode for a button to ground: A button connects pin 2 to ground when pressed, with no external resistor. Choose the pinMode: enter 1 for INPUT, 2 for OUTPUT, or 3 for INPUT_PULLUP.
Show worked solution
3: INPUT_PULLUP enables an internal pull-up so the open input has a defined HIGH state.
Open pull-up button: Pin 2 is INPUT_PULLUP and its button to ground is open. Enter 1 for HIGH or 0 for LOW.
Show worked solution
1: HIGH. The pull-up holds the open input high.
Closed pull-up button: Pin 2 is INPUT_PULLUP and its button is closed to ground. Enter 1 for HIGH or 0 for LOW.
Show worked solution
0: LOW. The closed switch pulls the input to ground.
From Digital and Analog Signals
Ultrasonic Echo signal: Classify the Echo output of an HC-SR04 module: enter 1 for a timed digital pulse or 2 for an analog voltage proportional to distance.
Show worked solution
1: timed digital pulse. Measure pulse duration to obtain range.
From Digital and Analog Signals
Phototransistor signal: A phototransistor drives a load resistor without a comparator or threshold stage. Enter 1 for an inherently digital output or 2 for an analog light response.
Show worked solution
2: analog response. A threshold circuit can turn that response into a digital signal.
From Digital and Analog Signals
Potentiometer wiper signal: A potentiometer has its outer terminals at 0 V and 5 V. Classify the wiper voltage: enter 1 for digital or 2 for analog.
Show worked solution
2: analog. The wiper voltage varies continuously with position in the ideal model.
From Milling Vises and Strap Clamps
Higher reading at the right end. The indicator body is on the operator side of the fixed jaw. You verified that moving the jaw toward the body increases the reading. The left mounting point is the pivot; only the right end is adjusted. A left-to-right sweep reads 0.00 mm at the left and +0.08 mm at the right. Which way should the first small correction move the right end?
- Toward the indicator.
- Away from the indicator.
- Along the jaw toward the left end.
Show reasoning
Option 2. The right end is closer to the indicator than the left. Move it away, then repeat the sweep and recheck after final tightening. The sign is defined by the response test, not a remembered clockwise dial rule.
From Milling Vises and Strap Clamps
Lower reading at the right end. The indicator body is on the operator side of the fixed jaw. You verified that moving the jaw toward the body increases the reading. The left mounting point is the pivot; only the right end is adjusted. The left reading is +0.02 mm and the right reading is −0.03 mm. Choose the correction.
- Move the right end toward the indicator.
- Move the right end away from the indicator.
- Accept it because neither reading is large.
Show reasoning
Option 1. Right minus left is −0.03 − 0.02 = −0.05 mm. The right end is farther from the indicator. Bring it toward the indicator and resweep. Alignment is judged against the specified job tolerance, which this problem does not supply.
From Milling Vises and Strap Clamps
Reversed indicator response. The left mounting point is the pivot. In this new indicator arrangement you verify that moving the jaw toward the body makes the reading DECREASE. Left reads 0.00 mm; right reads +0.04 mm. Which correction should move the right end?
- Away from the indicator body.
- No correction can be inferred.
- Toward the indicator body.
Show reasoning
Option 3. With this reversed response, the positive change means the right end is farther away. Move it toward the body. You have enough information because the response was demonstrated; reversing the contact arrangement invalidates a memorized sign rule.
From Milling Vises and Strap Clamps
A jump halfway across the jaw. The indicator body is on the operator side of the fixed jaw. You verified that moving the jaw toward the body increases the reading. The left mounting point is the pivot; only the right end is adjusted. The reading starts at 0.00 mm, rises abruptly to +0.12 mm at one small spot, returns to 0.00 mm and stays there to the far end. What is the best next action?
- Tap the right end away until the +0.12 mm spot reads zero.
- Inspect and clean the local contact surface and check indicator stability.
- Accept the setup without checking the middle.
Show reasoning
Option 2. A local jump is not the smooth trend expected from simple jaw rotation. Check for debris, damage or unstable contact, then repeat the entire sweep. Equal endpoints alone do not explain the jump.
From Milling Vises and Strap Clamps
Stud near the part. A strap receives P = 7,200 N. The part-contact-to-stud distance a is 18 mm; the stud-to-heel distance b is 54 mm. What downward force does the strap apply to the part? Assume an ideal weightless strap with only part and heel contacts and a vertical stud load. Round to the nearest newton.
Show reasoning
Fpart = P × b / (a + b) = 7,200 × 54 / (18 + 54) = 5,400 N. The heel receives 1,800 N. The two shares sum to 7,200 N. Using a in the numerator gives the heel share, not the requested part share.
From Milling Vises and Strap Clamps
Stud near the heel. A strap receives P = 7,200 N. The part-contact-to-stud distance a is 54 mm; the stud-to-heel distance b is 18 mm. What downward force does the strap apply to the part? Assume an ideal weightless strap with only part and heel contacts and a vertical stud load. Round to the nearest newton.
Show reasoning
Fpart = P × b / (a + b) = 7,200 × 18 / (54 + 18) = 1,800 N. The heel receives 5,400 N. The two shares sum to 7,200 N. Using a in the numerator gives the heel share, not the requested part share.
From Milling Vises and Strap Clamps
Unequal clamp arms. A strap receives P = 5,200 N. The part-contact-to-stud distance a is 27 mm; the stud-to-heel distance b is 63 mm. What downward force does the strap apply to the part? Assume an ideal weightless strap with only part and heel contacts and a vertical stud load. Round to the nearest newton.
Show reasoning
Fpart = P × b / (a + b) = 5,200 × 63 / (27 + 63) = 3,640 N. The heel receives 1,560 N. The two shares sum to 5,200 N. Using a in the numerator gives the heel share, not the requested part share.
From Milling Vises and Strap Clamps
Clamp-force rounding. A strap receives P = 4,100 N. The part-contact-to-stud distance a is 23 mm; the stud-to-heel distance b is 48 mm. What downward force does the strap apply to the part? Assume an ideal weightless strap with only part and heel contacts and a vertical stud load. Round to the nearest newton.
Show reasoning
Fpart = P × b / (a + b) = 4,100 × 48 / (23 + 48) = 2,772 N. The heel receives 1,328 N. The two shares sum to 4,100 N. Using a in the numerator gives the heel share, not the requested part share.
From End Mill Anatomy and Cutter Selection
Choose an aluminum-slot cutter. A rigid setup must cut a full-width slot in aluminum. All tools listed fit the feature and holder. Which is the strongest starting choice?
- A four-flute steel finishing tool with little chip space, selected only because it has more teeth.
- A two-flute aluminum-rated tool with open chip space and published slotting data.
- Any ball end mill, because end shape does not matter.
Show reasoning
Option 2. The material-specific geometry and slotting data address the actual cut. A suitable three-flute aluminum tool could also work, but it is not one of these choices. Tool material, reach, speed and feed still need checking.
From End Mill Anatomy and Cutter Selection
Choose a steel side-milling cutter. A rigid machine and fixture will side-mill steel. Each candidate has enough reach and fits the holder. Which starting choice is best justified?
- A steel-rated four-flute carbide tool with data for the intended engagement.
- An aluminum-specific tool chosen only for its polished appearance.
- A drill held in a drill chuck because its diameter matches.
Show reasoning
Option 1. It matches both material and side-milling duty. Four flutes and carbide are not guarantees by themselves; the steel rating, setup and application data are what make this choice justified.
From End Mill Anatomy and Cutter Selection
When HSS is a reasonable choice. A modest-speed machine will make a light interrupted cut. The fixture is secure, and the tool maker permits this operation with either listed milling tool. Which reasoning is sound?
- Carbide is always best because hardness alone determines the result.
- Choose a drill chuck to absorb the interruptions.
- A suitable HSS end mill may be economical and more tolerant of impact; use its lower-speed data.
Show reasoning
Option 3. Toughness, available speed and setup matter alongside hot hardness. This does not license a loose setup or justify using HSS outside its cutting-data range.
From End Mill Anatomy and Cutter Selection
Flutes do not prove plunge capability. A pocket needs an entry from above. You have a two-flute end mill, but its entry capability is unknown. What should decide whether to plunge?
- Two flutes automatically mean it can drill straight down.
- Confirm center-cutting geometry and the manufacturer’s permitted entry method and limits.
- Any end mill can plunge if the feed is slow enough.
Show reasoning
Option 2. Flute count alone does not establish center-cutting geometry. Choose a permitted ramp, plunge or prepared entry as the tool and operation require.
From Milling Chatter
Chatter after a longer tool. A formerly clean cut chatters after changing to a much longer end mill. Seating, runout and workholding have been checked. The feature does not need the extra reach. What is the first useful change?
- Shorten the unsupported tool reach while retaining the required shank engagement.
- Increase the overhang further.
- Change speed, feed and depth together so the cause is impossible to isolate.
Show reasoning
Option 1. The changed reach is a clear loss of stiffness. Correct it before tuning around it. Use the shortest reach that still clears the work and holder.
From Milling Chatter
Chatter only inside corners. A correctly seated, rigid setup cuts straight pocket walls cleanly but chatters in tight inside corners. What should you review first?
- Whether halving feed everywhere will hide the symptom without changing engagement.
- Whether doubling the axial depth will always damp the vibration.
- The increase in cutter engagement and the programmed corner path.
Show reasoning
Option 3. More of the circumference can become engaged in a corner. Smooth or reduce engagement using a suitable path and cutting data, then test the change.
From Milling Chatter
A stable setup at an unstable speed. Seating, rigidity and engagement have been reviewed. Chatter appears at the current speed. Which next experiment is useful?
- Always halve feed while leaving speed unchanged.
- Try a different permitted spindle speed and review feed to preserve the intended chip load.
- Assume that a higher speed can never be stable.
Show reasoning
Option 2. Speed changes the timing of tooth engagement. A different permitted speed may help, but the accompanying feed must still produce a suitable chip load. Slowing feed alone can cause rubbing.
From Milling Chatter
The part lifted during a cut. After ending a noisy cut, you find a gap between the part and one parallel. What takes priority before another trial?
- Raise spindle speed to move away from resonance.
- Lower feed until the sound is hidden.
- Correct seating and grip, then recheck location before cutting.
Show reasoning
Option 3. A moved part is a setup failure. Restore support and clamping and verify the part reference; do not try to tune speed around it.
From DC Motor Torque, Speed, and Power
Reflect load through a reduction: A 20-tooth motor gear drives a 40-tooth output gear through one mesh modeled at 90% efficiency. The output load torque is 0.600 N·m. What motor torque is required at a steady running point? Give four significant figures.
Show worked solution
The reduction is i = 40/20 = 2. With stated running mesh efficiency η = 0.90, Tmotor = Tload/(iη) = 0.600/(2 × 0.90) = 0.3333 N·m. Using 0.300 N·m ignores the modeled loss.
From DC Motor Torque, Speed, and Power
Find geared output speed: The motor has 0.800 N·m stall torque and 120 rpm no-load speed. It supplies the 0.3333 N·m reflected torque from the previous problem. Using the ideal linear motor curve and the 2:1 reduction, what is output speed?
Show worked solution
nmotor = 120(1 − 0.3333/0.800) = 70.0 rpm. The 2:1 reduction gives noutput = 70.0/2 = 35.0 rpm. Dividing the 120 rpm no-load endpoint by two would ignore the loaded motor curve.
From DC Motor Torque, Speed, and Power
Find gearbox output power: The motor operates at 0.3333 N·m and 70.0 rpm. Find motor shaft power, then apply the stated 90% running mesh efficiency. Give output power in watts. You may use π = 3.14 and four-significant-figure intermediate values.
Show worked solution
Pmotor = Tω = 0.3333 × 70.0 × 2π/60 = 2.443 W. Then Poutput = 0.90 × 2.443 = 2.199 W. Multiplying torque by rpm without converting to rad/s gives the wrong units and value.
From DC Motor Torque, Speed, and Power
Explain the comparison: In the lesson example, direct output power is 1.885 W while geared output power is 2.199 W. Which explanation matches the model? 1 = the gearbox created power; 2 = both cases had equal motor shaft power; 3 = reflected torque moved the motor to a different speed and shaft-power point before mesh loss; 4 = the mesh was more than 100% efficient.
Show worked solution
3. The reduction changes the motor torque required by the same output load. The motor therefore settles at a different speed and produces different shaft power. The stated mesh loss is then applied. This is not free power and does not claim equal electrical input.
From DC Motor Torque, Speed, and Power
Compare direct overload with geared running: Keep the 0.800 N·m, 120 rpm motor and the 2:1, 90% reduction, but raise output load torque to 0.900 N·m. Direct drive has no forward steady operating point in this model. What geared output speed does the model give?
Show worked solution
The reflected motor torque is 0.900/(2 × 0.90) = 0.500 N·m. The motor curve gives 120(1 − 0.500/0.800) = 45.0 rpm. After the 2:1 reduction, output speed is 22.5 rpm. This steady-running point does not prove the system starts or runs continuously.
From Selecting a Material-Removal Process
A small-radius die opening
A hardened steel die needs a through opening with small internal radii. A start hole is available. Which candidate best fits?
- Wire EDM
- Pure waterjet
- Ordinary drilling
Show answer and reasoning
Choice 1. Wire EDM can trace the conductive hardened part with low cutting force. Check the required radius against wire radius plus gap; perfectly sharp corners are not possible.
From Laser, Waterjet and Plasma Cutting
A thick glass profile
A thick, non-tempered glass plate needs a through profile without a thermal HAZ. Which candidate fits?
- Plasma
- Abrasive waterjet
- Wire EDM
Show answer and reasoning
Choice 2. Abrasive waterjet can cut nonconductive glass without a thermal HAZ. Thickness, piercing and edge quality still require checks.
From Electrical Discharge Machining
A nonconductive ceramic
Why is ordinary EDM unsuitable for an electrically insulating ceramic?
- The ceramic is too hard for every machining process
- EDM requires electrical conduction through the work
- EDM only makes round holes
Show answer and reasoning
Choice 2. EDM needs a discharge circuit through the workpiece. Hardness alone is not the deciding property.
From Electrical Discharge Machining
A blind die cavity
Which is the stronger EDM candidate for a blind cavity in hardened conductive steel?
- Wire EDM threaded through the cavity floor
- Sinker EDM with a shaped electrode
- Neither, because hardened steel cannot be EDMed
Show answer and reasoning
Choice 2. A sinker electrode can approach a blind cavity. Wire EDM requires a through path for the wire.
From Electrical Discharge Machining
EDM removal energy
What removes material at the work in EDM?
- A sharp edge shearing chips
- Heat from electrical discharges
- Electrochemical dissolution without sparks
Show answer and reasoning
Choice 2. Electrical power produces discharges whose local heat melts and vaporizes material.
From Electrochemical Machining
ECM removal mechanism
Which mechanism describes ECM?
- Anodic dissolution through a conductive electrolyte
- Laser heating through insulating oil
- Abrasive impacts
Show answer and reasoning
Choice 1. The work is the anode. Current through the electrolyte supports dissolution; ECM is not spark erosion.
From Laser, Waterjet and Plasma Cutting
Laser removal energy
Which supplies the concentrated cutting energy in laser cutting?
- Moving abrasive grit
- An arc through a wire electrode
- Focused light
Show answer and reasoning
Choice 3. The beam supplies thermal energy. The machine and wavelength must suit the material.
From Laser, Waterjet and Plasma Cutting
Plasma removal energy
Which describes ordinary plasma cutting?
- An electric arc in ionized gas melts conductive work
- A chemical bath dissolves glass
- Water alone cuts a hard steel die
Show answer and reasoning
Choice 1. The arc supplies heat and the gas carries molten material away. Conductive metal is the candidate material.
From Laser, Waterjet and Plasma Cutting
Waterjet removal energy
What supplies removal energy in abrasive waterjet cutting?
- Electrical discharges through the work
- Mechanical energy of moving water and grit
- Light absorbed by the work
Show answer and reasoning
Choice 2. The jet carries particles that erode material. Electricity may power the pump, but the removal mechanism is mechanical.
From Ultrasonic Machining
Ultrasonic removal energy
In ultrasonic impact machining, what acts on the hard, brittle work?
- A shaped tool driving abrasive particles by vibration
- An ionized gas arc
- Electroplating
Show answer and reasoning
Choice 1. The vibrating tool drives abrasive impacts through a slurry. This is distinct from adding vibration to conventional cutting.
From Chemical Milling and Etching
Chemical removal mechanism
What removes exposed metal in ordinary chemical milling?
- A laser
- A rotating tool
- Reaction with an etchant
Show answer and reasoning
Choice 3. A mask protects selected areas while the etchant dissolves exposed metal. There is no external machining current as in ECM.
From Laser, Waterjet and Plasma Cutting
A soft foam gasket
Which is a reasonable first candidate for a soft foam gasket?
- Pure waterjet
- Plasma
- Wire EDM
Show answer and reasoning
Choice 1. Soft material may be separated with water alone. Abrasive is useful for harder materials; the gasket does not need to conduct electricity.
From Selecting a Material-Removal Process
Explain a no-HAZ choice
A conductive plate must retain an edge without thermal alteration. Explain why abrasive waterjet is a candidate and why EDM is not automatically a substitute.
Self-check: write first, then compare. Your response is not automatically graded or saved.
Show a model answer
Abrasive waterjet removes by mechanical erosion and normally produces no thermal HAZ. EDM removes by local melting and vaporization, so recast and thermal surface effects remain possible despite low cutting force. Waterjet edge quality and mechanical damage still need evaluation.
From Electrochemical Machining
Distinguish the two liquids
Explain the difference between ECM electrolyte and EDM dielectric, including why “ECM makes no heat” is misleading.
Self-check: write first, then compare. Your response is not automatically graded or saved.
Show a model answer
ECM electrolyte conducts machining current and carries reaction products. EDM dielectric normally insulates until local breakdown permits a discharge, then helps cool and flush the gap. ECM does not melt the work, but current heats the electrolyte.
From Laser, Waterjet and Plasma Cutting
Thickness and finish
A waterjet profile has acceptable dimensions at the top but an unacceptable lower edge. Name two process effects to investigate.
Self-check: write first, then compare. Your response is not automatically graded or saved.
Show a model answer
Check taper (different cut widths through the thickness) and jet lag or striations. Thickness, travel speed and the cutting setup affect both. A no-HAZ process does not guarantee straight walls or the required surface finish.
From Chemical Milling and Etching
Etching a narrow land
Why might a narrow strip of protected metal end up narrower than its mask pattern?
Self-check: write first, then compare. Your response is not automatically graded or saved.
Show a model answer
The etchant can attack sideways beneath the mask edge as well as downward. This undercut must be allowed for in mask design and limits small features, especially for deeper etches.
From Electrical Discharge Machining
Wire center-path offset
For this teaching example, the wire radius is 0.125 mm and the one-sided spark gap is 0.025 mm. What is the offset magnitude from the finished wall to the wire center path? Give mm rounded to the nearest 0.001 mm.
Show answer and reasoning
Offset = radius + gap = 0.125 + 0.025 = 0.150 mm. The center path is on the removed-material side of the wall. 0.125 mm ignores the gap; 0.175 mm adds two gaps; 0.275 mm uses the wire diameter. The full kerf is 0.300 mm, not the one-sided offset.
From Extrude and Revolve
Choose a base feature. A round knob has several coaxial outside steps and a central bore. Which base best keeps those radial and axial dimensions in one section? Choose 1 = revolve, 2 = shell, 3 = linear pattern.
Show reasoning
1: Revolve. A section can describe the steps and bore together. This is a design-intent choice; extrusions could also construct the shape.
From Extrude and Revolve
Keep a cut through. A hole must pass through a single plate even when the plate thickness increases. Choose 1 = blind at the current thickness, 2 = through all in the cutting direction, 3 = blind at half the thickness.
Show reasoning
2: Through all expresses the requirement. A blind cut tied only to the old thickness may leave a floor.
From Extrude and Revolve
Split a total depth. A SOLIDWORKS Mid Plane extrusion has a total depth of 18 mm. How far does it extend from the sketch plane on each side?
Show reasoning
18 ÷ 2 = 9.00 mm on each side. Using 18 mm on each side would double the total depth.
From Extrude and Revolve
Count a minimum with stated limits. Starting with no solid, make a plain rectangular block with one circular through hole. Allow only ordinary straight boss extrusions from closed sketch regions, including nested loops. Count solid features only, not sketches. What is the minimum number of solid features?
Show reasoning
1 feature. Extrude a rectangular region with a circular region excluded all the way through. A separate boss and cut uses 2 features and may express independent design intent better, but is not the minimum under these rules.
Find a lost reference. The tree is: plate extrusion; hole using a sketch on its top face; pattern using that hole; an unrelated edge chamfer. Delete only the hole while retaining the plate. Which feature definitely loses its stated input? Choose 1 = plate extrusion, 2 = hole pattern, 3 = every earlier feature.
Show reasoning
2: The hole pattern loses its seed. Deletion dialogs may offer dependent deletions. Do not infer that every later feature must fail merely because it appears later.
Estimate mass from volume. A uniform solid spacer is 24 × 15 × 10 mm with no holes. Use the assumed density 2.50 g/cm³. What is its mass in grams? Give two decimal places.
Show reasoning
Volume = 24 × 15 × 10 = 3,600 mm³ = 3.6 cm³. Mass = 3.6 × 2.50 = 9.00 g. There are 1,000 mm³ in 1 cm³; multiplying 3,600 directly by the density would give a result 1,000 times too large.
From Sweeps and Lofts
Recognize a profile and path. You have a circular section and a smooth bent centerline. You want a solid rod of constant circular size along that line. Choose the simplest starting feature: 1 = chamfer, 2 = sweep, 3 = shell.
Show reasoning
2: Sweep the circular profile along the path. Verify profile alignment and enough bend radius to avoid self-intersection.
From Sweeps and Lofts
Connect changing sections. You have a large oval section, a medium oval section, and a small oval section on three separated planes. Which feature directly transitions through these ordered sections? Choose 1 = linear pattern, 2 = fillet, 3 = loft.
Show reasoning
3: Loft. Select the profiles in spatial order and inspect the connections. Add guides only when needed to control the shape between sections.
From Patterns, Mirror, and Applied Features
Count the spaces in a row. A single-direction linear pattern has 5 total hole instances including the seed, with 12 mm between adjacent centers. No instances are skipped. What is the first-to-last center distance?
Show reasoning
There are 5 − 1 = 4 intervals. Span = 4 × 12 = 48.00 mm. Multiplying by 5 counts one interval too many.
From Patterns, Mirror, and Applied Features
Space a full-circle pattern. Eight total holes, including the seed, are evenly spaced around a complete circle, with no duplicate at 360°. What is the angular pitch between adjacent holes?
Show reasoning
360° ÷ 8 = 45.00°. The last-to-first gap is also an interval; dividing by 7 is for a different endpoint-inclusive arc construction.
From Patterns, Mirror, and Applied Features
Choose symmetry. A tab must have an oppositely handed counterpart at equal distance across a center plane. Choose 1 = mirror, 2 = translate with a linear pattern, 3 = shell.
Show reasoning
1: Mirror reflects the tab. Translation preserves handedness and does not encode plane symmetry.
From Patterns, Mirror, and Applied Features
Hollow a solid. A solid block must become an open container with a controlled wall and floor thickness. Choose 1 = fillet, 2 = shell with the top face removed, 3 = circular pattern.
Show reasoning
2: Shell. Select the opening face, thickness, and inward or outward direction, then inspect the result in section.
From Patterns, Mirror, and Applied Features
Add a support web. You have a sketch line spanning two supporting walls and need a thin reinforcing web with controlled thickness. Choose 1 = chamfer, 2 = shell, 3 = rib.
Show reasoning
3: Rib. The sketch, thickness, direction, and target part define the support. Check that the web joins the walls.
From Patterns, Mirror, and Applied Features
Tilt faces for release. Side faces must taper relative to a pull direction while the neutral-plane section stays fixed. Choose 1 = draft, 2 = mirror, 3 = circular pattern.
Show reasoning
1: Draft. Check which end grows or shrinks. The required angle depends on the design and process; it is not supplied by this question.
From Patterns, Mirror, and Applied Features
Choose a cylindrical recess. A hole needs a larger cylindrical recess with a flat shoulder above its smaller bore. Choose 1 = countersink, 2 = counterbore, 3 = plain hole with no recess.
Show reasoning
2: Counterbore. A countersink has a conical recess. Define the type and size separately from the center position.
From Patterns, Mirror, and Applied Features
Distinguish bevel from radius. A sharp outside edge must become a straight bevel in section, rather than a rounded transition. Choose 1 = chamfer, 2 = fillet, 3 = rib.
Show reasoning
1: Chamfer. A fillet gives a curved transition; a rib adds a support web.
From Pictorial Sketching
Identify the oblique pictorial. In the comparison, which numbered panel uses oblique sketching?
Show explanation
Panel 2 preserves the rectangular front face and sends parallel edges along one receding direction. Panel 1 uses isometric directions.
From Pictorial Sketching
Locate a non-isometric line. Which line on the beveled block is non-isometric?
Show explanation
PQ is parallel to none of the three isometric axes. Locate P and Q using axis-parallel offsets, then join them.
From Pictorial Sketching
Sketch a circle on an isometric top face. A circular feature lies on the horizontal top face of an isometric block. What should its sketch look like?
Show explanation
The face is tilted to the picture plane, so the circle appears elliptical. The four midpoints locate tangencies; joining them with straight lines does not form the curve.
From Visualizing Parts from Multiview Drawings
Find the edge views of face A. In the surface diagram, which principal views show face A as an edge?
Show explanation
A is parallel to the front viewing plane, so front shows its true shape. Top and right look along that face and collapse it to a line.
From Visualizing Parts from Multiview Drawings
Match a beveled block to its top view. Which candidate is the top view? The front of the part is at the bottom of every candidate.
Show explanation
Candidate 2. The bevel runs through the entire depth and begins two-thirds of the way across the width from the left. Its boundary therefore runs front to back near the right side. Candidate 1 changes along depth; candidate 3 puts the boundary on the wrong side.
From Visualizing Parts from Multiview Drawings
Classify sloping face B. B appears as an edge in front and a foreshortened area in top and right. What type of flat surface is it?
Show explanation
Inclined. B is perpendicular to the front projection plane and tilted to the other two. The sloping line in front is its edge view.
From Visualizing Parts from Multiview Drawings
Recognize an oblique surface. A flat face is tilted to all three principal planes, parallel to none and perpendicular to none. Which statement is correct?
Show explanation
An oblique surface shows an area in all three principal views, but none is its true shape. This describes surface orientation, not the pictorial method.
From Visualizing Parts from Multiview Drawings
Check before extruding a profile. You have recognized the front profile. What must you check before carrying it unchanged through the full depth?
Show explanation
Top and side must support a constant cross-section. A pocket or step may end partway through the depth; the front outline alone cannot establish that. A matching overall width and height confirms only the enclosing box.
Monomers and polymers: Which statement correctly connects monomers and polymers?
- A monomer is a finished plastic part.
- Monomers react to build polymer molecules.
- Polymer means any small metal particle.
Show answer and reasoning
Choice 2. Small starting molecules react to build the much larger polymer molecules. Polymer chains may also branch or become chemically linked.
Reheating cured epoxy: A fully cured epoxy casting must be reshaped. Which prediction fits its material type?
- It is a thermoplastic and can simply be remelted.
- It is a thermoset and cannot be remelted into a new shape.
- It becomes a monomer automatically when warmed.
Show answer and reasoning
Choice 2. Cured epoxy is a thermoset. Its chemically linked network does not become a reusable melt. Heating is not a way to return it to the uncured liquid.
Reheating polypropylene: A clean PP part can soften and flow again under appropriate processing conditions. What does that tell you?
- It is a thermoplastic; local recycling acceptance is a separate question.
- It is a thermoset because it started as a solid.
- Every local recycling program must accept it.
Show answer and reasoning
Choice 1. PP is a thermoplastic. Remeltability is useful for recycling, but the actual item still needs a compatible collection and processing route.
From Choosing a Plastic Forming Process
A long run of clips: A manufacturer needs 250,000 identical PP clips with small ribs and a controlled mating shape. The design can release from a mold. Which process is the strongest starting candidate?
- Hand layup
- Rotational molding
- Injection molding
- Extrusion
Show answer and reasoning
Choice 3: Injection molding. Injection molding repeats detailed discrete parts. The quantity can spread the mold cost. Extrusion cannot directly vary the shape along each clip.
From Choosing a Plastic Forming Process
Thin-walled bottles: A packaging line needs 600,000 hollow HDPE detergent bottles with narrow necks. Which process best fits the shape and quantity?
- Blow molding
- Compression molding
- Hand layup
- Casting
Show answer and reasoning
Choice 1: Blow molding. Blow molding expands hot plastic into hollow containers. HDPE bottles commonly use extrusion blow molding. It does not need a removable core filling the entire bottle interior.
From Choosing a Plastic Forming Process
Large hollow tanks: An order calls for 240 large hollow PE water-storage tanks. Fine surface detail is unnecessary and longer molding cycles are acceptable. Which process deserves the first investigation?
- Extrusion
- Injection molding
- Single-sheet vacuum forming
- Rotational molding
Show answer and reasoning
Choice 4: Rotational molding. Rotational molding suits large hollow PE shells and modest quantities. Its slower heating and cooling cycle is acceptable here. Confirm wall control and delivery rate before final selection.
From Choosing a Plastic Forming Process
A short run of trays: A supplier needs 180 shallow PETG trays, each open on top, using available sheet. Which process fits the starting stock and geometry?
- Blow molding
- Vacuum forming
- Rotational molding
- Compression molding of thermoset compound
Show answer and reasoning
Choice 2: Vacuum forming. Vacuum forming reshapes hot thermoplastic sheet against a tool. A simple tool can suit a short run. Check draw depth, thinning and draft.
From Choosing a Plastic Forming Process
Continuous channel: A building product needs thousands of meters of PVC channel with the same cross-section throughout. Which process fits?
- Extrusion
- Casting
- Hand layup
- Blow molding
Show answer and reasoning
Choice 1: Extrusion. Extrusion produces a continuous profile through a die. End features or isolated side holes would need later operations.
From Choosing a Plastic Forming Process
Repeated thermoset parts: A producer needs 40,000 electrical components from a thermosetting molding compound. Both faces need shaped detail, and the charge can spread in matched heated tools. Which process fits?
- Single-sheet vacuum forming
- Rotational molding
- Compression molding
- Extrusion of thermoplastic pipe
Show answer and reasoning
Choice 3: Compression molding. Compression molding presses the charge between matched tools and allows the thermoset to cure. A cooling-only explanation would miss the chemical change.
From Choosing a Plastic Forming Process
Twenty detailed replicas: A designer needs 20 small detailed solid replicas in a casting-grade two-part polyurethane. Flexible molds are available and cure time is acceptable. Which process fits?
- Extrusion
- Blow molding
- Vacuum forming
- Casting
Show answer and reasoning
Choice 4: Casting. Casting matches the liquid thermosetting resin and short run. Flexible tooling can reproduce detail, but mold life and trapped air still matter.
From Choosing a Plastic Forming Process
Custom reinforced shells: A maker needs eight broad glass-fiber/epoxy equipment covers. One smooth tool-side face is sufficient and manual work is acceptable. Which process fits?
- Blow molding
- Wet hand layup
- Extrusion
- Injection molding of unfilled PP
Show answer and reasoning
Choice 2: Wet hand layup. Wet hand layup builds reinforcement and thermosetting resin against a mold. The resulting material belongs to Wood & Composites in the course model. A one-sided mold does not precisely control both faces.
From Plastic Forming Processes
A dimple opposite a thick rib: An injection-molded cover has a shallow depression on the visible face directly opposite a heavy rib. Which design change should be investigated first?
- Make every rib as thick as possible.
- Reduce the heavy rib or core out the local mass.
- Remove all draft so the part stays in place.
Show answer and reasoning
Choice 2. The thick region can shrink differently and pull the surface inward, producing sink. Reduce the local mass and check packing and cooling; filling every region thicker is not a general cure.
From Plastic Forming Processes
A part scrapes during ejection: A deep molded wall is parallel to the opening direction and scrapes the tool as the part is pushed out. Which feature is missing or insufficient?
- Draft
- Resin identification code
- A higher cavity count
Show answer and reasoning
Choice 1: draft. A slight release taper reduces contact as the mold opens. Required draft depends on depth, texture, material and tooling.
From Plastic Forming Processes
A sudden thick-to-thin wall change: A housing changes abruptly from a thin wall to a bulky solid region. Which redesign best addresses the geometry problem?
- Make the bulky region larger.
- Add a recycling code to the thick region.
- Core out unnecessary material and use gradual transitions.
Show answer and reasoning
Choice 3. A more uniform section and gradual transitions help manage filling and shrinkage. The goal is a manufacturable shape, not merely a larger mass of plastic.
PET bottle resin: Which resin identification code matches PET (also called PETE)? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
1. Code 1 identifies polyethylene terephthalate. Do not generalize that code to every material with PET in its name.
HDPE container resin: Which resin identification code matches HDPE? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
2. Code 2 identifies high-density polyethylene.
PVC resin: Which resin identification code matches PVC (vinyl)? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
3. Code 3 identifies polyvinyl chloride.
LDPE film resin: Which resin identification code matches LDPE? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
4. Code 4 identifies low-density polyethylene.
PP lid resin: Which resin identification code matches PP? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
5. Code 5 identifies polypropylene. A code does not guarantee that a local program accepts that lid.
Polystyrene resin: Which resin identification code matches PS? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
6. Code 6 identifies polystyrene, including familiar rigid and foamed products.
The OTHER resin category: Which resin identification code matches a resin outside categories 1–6, such as polycarbonate? Enter a number from 1 to 7. This question identifies resin, not local recycling acceptance.
Show answer and reasoning
7. Code 7 is OTHER. It does not identify one resin, and it does not mean all code-7 items can be recycled together.
From Choosing a Plastic Forming Process
Parts per hour from four cavities: An ideal molding cycle takes 48 seconds and produces one part in each of four identical cavities. Assume no downtime or rejects. How many parts are produced per hour?
Report to the nearest whole part per hour.
Show answer and reasoning
3600 s/h ÷ 48 s/cycle = 75 cycles/h. Then 75 × 4 = 300 parts/h. Reporting 75 ignores the four cavities. Multiplying cycle time by cavity count does not give an hourly rate.
From Choosing a Plastic Forming Process
When two routes cost the same: Two hypothetical routes meet the same specification. A costs $1,200 for tooling plus $9 per part. B costs $13,200 for tooling plus $3 per part. Assume constant variable costs, no scrap and no tooling replacement. At what quantity do total costs match?
Report to the nearest whole part.
Show answer and reasoning
Set 1200 + 9N = 13200 + 3N. Thus 6N = 12000, and N = 2,000 parts. Each route then costs $19,200. Using $13,200 without subtracting A's tooling cost gives 2,200, which is wrong.
From Cutting and Machining Plastics
A melted routed edge. An acrylic slot has softened chips stuck to its edge. The tool is rubbing at a very low feed per tooth. Which is the best next investigation?
Show reasoning
Choice 2. Rubbing and trapped chips generate heat. A lower feed by itself can worsen the problem; the tool, feed, RPM and chip path need to work together.
From Cutting and Machining Plastics
Cutting a clear polycarbonate profile. A clear polycarbonate sheet needs a curved profile with a clean edge. Which is the best starting plan?
Show reasoning
Choice 3. Polycarbonate can discolor and char under CO2 cutting. A clear appearance does not establish either the resin or its processing behavior.
From Cutting and Machining Plastics
Preserving the molded wall. A molded part has a feed connection still attached at its gate. Which result should degating achieve?
Show reasoning
Choice 1. The connection is excess material, but the wall is part of the finished product. A notch can weaken it.
From Joining Plastics
A permanent clear corner. Two identified acrylic pieces have closely fitting edges and need a permanent clear corner. Which candidate deserves evaluation first?
Show reasoning
Choice 2. Compatible solvent cement can soften the mating surfaces and develop a joint as solvent leaves. Fit, residual stress and the full strength-development time still matter.
From Joining Plastics
An adhesive peels from polypropylene. A fully cured general-purpose adhesive peels cleanly from a clean PP sample. What material issue should be investigated?
Show reasoning
Choice 3. PP resists wetting by many ordinary adhesives. Specialized systems can work, so neither impossible nor just add more glue is a sound conclusion.
From Joining Plastics
A cover that opens repeatedly. A thermoplastic cover must be removed regularly for inspection. Which candidate best addresses that requirement?
Show reasoning
Choice 1. A mechanical joint permits disassembly. Insert geometry and the plastic supporting the insert still govern pullout and cracking; a metal thread alone is not enough.
From Joining Plastics
A weld that looks continuous. An extrusion weld looks smooth but leaks during the specified inspection. Which conclusion is justified?
Show reasoning
Choice 2. Compatible filler must bond to adequately prepared and heated base material. A continuous-looking bead is not proof of a leak-tight joint.
From Joining Plastics
A joint creeps in warmth. A part bonded with an ordinary hot-melt adhesive holds at room temperature but slips under sustained load when warm. What is the likely selection issue?
Show reasoning
Choice 3. Hot-melt adhesive forms its own bond layer. Fast handling strength does not establish resistance to a sustained load at a higher temperature.
From FDM 3D Printing
A printed bar in tension. A rectangular FDM test bar will be pulled along its length. With comparable material and settings, which orientation is the better starting candidate?
Show reasoning
Choice 1. Deposited roads can carry the main load along the layers. Across-layer tension relies on interlayer bonds. This is a starting choice to test, not a guaranteed strength value.
From FDM 3D Printing
An inaccessible supported roof. The slicer places ordinary breakaway support under a long roof inside a closed cavity. Which change should be evaluated?
Show reasoning
Choice 2. Support must be removable. Reorientation, splitting the part for assembly or redesigning the roof may help; finer layers alone do not solve an inaccessible support.
From Conditioning and Finishing Plastics
Painting a detailed PLA part. A detailed PLA part needs an opaque painted finish while preserving fit. Which candidate is most appropriate to trial?
Show reasoning
Choice 3. Mechanical preparation and a material-compatible coating are appropriate candidates. ABS acetone smoothing and acrylic flame polishing are not transferable recipes for PLA.
From Conditioning and Finishing Plastics
A print changes size after annealing. A PLA part has better heat resistance after a validated annealing trial, but no longer fits its mating part. What should be concluded?
Show reasoning
Choice 1. Annealing can change crystalline structure and dimensions. Its usefulness depends on the complete requirement, including the fit and the grade-specific schedule.
From FDM 3D Printing
Count equal-height layers. A simplified print is 30 mm tall. Every layer, including the first, is 0.20 mm thick. How many layers does it have? Give a whole number.
Show worked solution
30 mm ÷ 0.20 mm/layer = 150 layers. This assumes constant layer height and says nothing about travel or heating time.
From FDM 3D Printing
Estimate deposited filament mass. Use a simplified volume model: walls and skins total 6 cm³, and the remaining interior is 30 cm³ at 20% infill. Assume a density of 1.24 g/cm³. Ignore supports, purge, brim and other waste. Estimate mass in grams to the nearest 0.1 g.
Show worked solution
Deposited volume = 6 + (30 × 0.20) = 12 cm³. Mass = 12 × 1.24 = 14.88 g, or 14.9 g to the nearest 0.1 g. Applying 20% to the entire 36 cm³ would incorrectly reduce the solid walls and give 8.928 g. Real slicer paths need not match this ideal volume model exactly.
Phase-locked PWM sample. An ideal 0–5 V PWM waveform is HIGH for normalized phase [0, 0.25) and LOW for [0.25, 1). A point sampler runs once per PWM period, always at phase 0.10. What does every sample read? 1 = 5 V; 2 = 1.25 V; 3 = 0 V; 4 = it alternates.
Show worked solution
1. Phase 0.10 lies inside the HIGH interval, so every point sample is 5 V. The complete-cycle time mean is still 25% × 5 V = 1.25 V; the sampler is not averaging a cycle.
LOW-phase finite sample mean. An ideal 25%, 0–5 V waveform is HIGH on normalized phase [0, 0.25) and LOW on [0.25, 1). It is sampled once per PWM period at phase 0.60 for 20 point samples. What is the mean of those selected samples?
Show worked solution
Phase 0.60 is LOW on every cycle, so all 20 samples are 0 V and their mean is 0 V. A larger count does not remove phase locking.
Four-phase PWM sample mean. An ideal 25%, 0–5 V waveform is HIGH on normalized phase [0, 0.25) and LOW on [0.25, 1). One repeating four-sample lattice lands at phases 0.01, 0.26, 0.51 and 0.76. What is the analog mean of those four point values?
Show worked solution
Only phase 0.01 is HIGH: (5 + 0 + 0 + 0)/4 = 1.25 V. This chosen lattice matches the complete-cycle time mean; faster sampling by itself does not guarantee that result.
From Analog-to-Digital Conversion
Mean of ADC lower edges. In the ideal 10-bit, 5.000 V bin model, the four raw samples 5, 0, 0 and 0 V convert to codes 1023, 0, 0 and 0. Code 1023 reconstructs to lower edge 4.9951171875 V. What is the mean of the four reconstructed lower edges?
Show worked solution
(4.9951171875 + 0 + 0 + 0)/4 = 1.248779296875 V. That is slightly below the 1.25 V analog sample mean because each code represents its bin's lower edge.
Settled RC ripple below the mean. The lesson's ideal unloaded RC model uses 25% duty, 500 Hz and τ = 10 ms. Its complete-cycle mean is 1.25 V; after settling, the phase-0 point is 1.157869414 V. How far below the cycle mean is that point, to four decimal places?
Show worked solution
1.25 − 1.157869414 = 0.092130586 V, or 0.0921 V. The phase-0 sample is instantaneous settled ripple, not the cycle mean, and this model does not simulate startup.
Recognize a visible edge. Which line would you read as an edge directly visible from this viewing direction?
Show explanation
Thick continuous line. A relatively thick continuous line is a visible or object line.
Distinguish an axis from an edge. A thin long–short line passes through the center of a circular hole. What does it identify?
Show explanation
The hole center or axis. The center line locates the hole center or axis. It is not a physical edge.
Recognize an alternate position. A lever outline is repeated in a thin long–short–short pattern beside its solid outline. What is the likely meaning?
Show explanation
An alternate lever position. Phantom lines can show an alternate position. Use nearby notes to confirm what the alternate outline represents.
Read a dimension line. A thin line has arrowheads at both ends and the value 42 between them. What kind of line is it?
Show explanation
Dimension. The dimension line identifies the extent of the stated measurement. Read the value and the drawing units rather than measuring the image.
Read an extension line. Thin lines start just outside two part edges and extend to the ends of a dimension line. What are they?
Show explanation
Extension lines. Extension lines carry feature locations to the dimension line. The small gap helps distinguish them from part edges.
Follow a note to its feature. A thin bent line joins a hole note to an arrow touching the hole outline. What is the line?
Show explanation
Leader. A leader connects the note to the feature it describes. Follow its endpoint before applying the note.
Read cutting-plane arrows. A prominent line across a view has arrows and the letter B at both ends. A separate view is labeled SECTION B–B. What do the arrows tell you?
Show explanation
The direction to look into the section. The arrows set the viewing direction for the imaginary cut. B–B connects the cut location to its section view.
Read section hatching. Thin diagonal strokes fill a region in a section view. What do they ordinarily show?
Show explanation
Material crossed by the imaginary cut. Hatching identifies cut material. It does not instruct someone to machine diagonal grooves.
Read a shortened view. A long uniform bar is drawn with zigzag break lines and an overall length dimension. How should you interpret it?
Show explanation
Part of the view is omitted; read the full length dimension. A break can shorten a long view on the sheet. It does not by itself mean the physical bar is fractured.
Resolve coincident line types. A visible edge and a hidden edge project to exactly the same segment. Which line should you expect there?
Show explanation
Visible line. The visible line takes precedence over the hidden line at the overlap. Consult another view to recover the concealed geometry.
From Orthographic Projection and the Glass Box
Identify the front view. Which candidate is the front view of the pictured part? The numbered candidates are laid out as choices, not as a projection arrangement.
Show explanation
View 2. View 2 follows the tall-left stepped profile. Width is horizontal and height is vertical.
From Orthographic Projection and the Glass Box
Identify the top view. Which candidate is the top view of the pictured part? The numbered candidates are laid out as choices, not as a projection arrangement.
Show explanation
View 3. View 3 shows the rectangular footprint, split at the width where the height changes. Looking down removes height from the outline.
From Orthographic Projection and the Glass Box
Identify the right-side view. Which candidate is the right view of the pictured part? The numbered candidates are laid out as choices, not as a projection arrangement.
Show explanation
View 1. View 1 shows depth and height. The near low shelf gives the visible horizontal edge; the tall strip behind it reaches the top outline.
From Orthographic Projection and the Glass Box
Recognize a face seen edge-on. You look exactly along a flat face so that the face is edge-on to you. How does that face project?
Show explanation
As a line. An edge-on face collapses to a line. Its full area appears only when the viewing direction exposes the face.
Decide whether a plate needs another view. A flat plate has a fully specified face outline and holes. A note states its uniform thickness; there are no other features. Must you add two more views just to reach three?
Show explanation
No, a complete one-view description can be enough. The target is an unambiguous definition. Under the stated assumptions, a face view plus thickness information can define the shape.
Find information missing from one view. Two solid blocks have identical front outlines but different depths. Which added view directly distinguishes their depths?
Show explanation
A top view at the same scale. The top view shows depth. Repeating or enlarging the front view does not reveal it.
From First-Angle and Third-Angle Projection
Place a right view in first-angle. The drawing declares first-angle projection. Where is the right-side view normally placed relative to the front?
Show explanation
To the left. First-angle puts the right-side view to the left of the front. The name still identifies the direction from which you look.
From First-Angle and Third-Angle Projection
Place a top view in third-angle. The drawing declares third-angle projection. Where is the top view normally placed relative to the front?
Show explanation
Above. Third-angle places the top above the front. Front and top remain aligned by width.
From First-Angle and Third-Angle Projection
Confirm the drawing convention. A drawing from a U.S. supplier has a clearly marked first-angle symbol. What should you do when interpreting the layout?
Show explanation
Use the declared first-angle convention and check view consistency. The explicit convention controls interpretation. A country or unit system is not a substitute for the symbol.
From Angular Motion
1. Index table in radians: A rotary index table advances 40° between one station and the next. How many radians is one advance?
Show worked solution
Put degrees on the bottom so they cancel:
40° × = 0.6981317… rad
or 0.6981 rad. The same thing exactly: 40° × π/180° = 2π/9 rad. A quick check on the size of the answer — one radian is about 57.3°, so 40° should come out a little under 1 rad, and it does.
From Angular Motion
2. Nameplate speed into the formula: A gearbox output shaft turns at 340 r/min. What is its angular velocity in rad/s?
Show worked solution
340 r/min × = 35.60472… rad/s
or 35.60 rad/s. The factor is not measured, it is the two definitions divided: 2π rad per revolution ÷ 60 s per minute = 0.104719755…
From Angular Motion
3. Spindle slowing down: A spindle is running at 1,800 r/min. The drive brings it down to 600 r/min in 4.0 s. What is its average angular acceleration in rad/s2, taking the running direction as positive?
Show worked solution
Only the change in speed matters, so take it first:
Δω = 600 r/min − 1,800 r/min = −1,200 r/min
−1,200 r/min × = −125.6637 rad/s
Then α = Δω / Δt = −125.6637 rad/s ÷ 4.0 s = −31.4159 rad/s2, or −31.42 rad/s². The sign is the answer as much as the number is: it says the angular acceleration points opposite the direction of rotation, which is what slowing down means. This is the average over the 4.0 s; a drive that ramps its braking has a changing instantaneous value inside the same interval.
From Angular Motion
4. Back to the nameplate: A calculation has produced an angular velocity of 52 rad/s, and the machine has to be set in r/min. What speed is that?
Show worked solution
Turn the factor over so that rad/s cancels:
52 rad/s × = 496.5634… r/min
or 497 r/min. Carry the factor to at least five figures — the exact value is 60/(2π) = 9.5492966…, and rounding it to 9.54 puts the answer about 0.1% low, which is a real error on a speed that has to be dialled in.
From Rotation and Linear Speed
5. Belt speed from a drive roller: A conveyor drive roller is 160 mm in diameter and turns at 60 r/min. What is the belt speed in m/s?
Show worked solution
- Halve the diameter:
r = 80 mm = 0.080 m. - Convert the speed:
60 r/min × 0.1047198 rad/s per r/min = 6.283185 rad/s. v = ωr = 6.283185 rad/s × 0.080 m = 0.5026548 m/s.
The belt runs at 0.503 m/s. Worth a sanity check: 60 r/min is one turn per second, and one turn of a 160 mm roller lays out π × 0.160 m = 0.5027 m of belt — the same number, reached without the formula. The answer assumes the belt does not slip on the roller.
From Rotation and Linear Speed
6. Setting the roller speed: Material has to feed past a station at 1.25 m/s. The drive roller is 250 mm in diameter. What speed must the roller be set to, in r/min?
Show worked solution
r = 125 mm = 0.125 m.- Rearrange:
ω = v/r = 1.25 m/s ÷ 0.125 m = 10 rad/s. - The machine is set in r/min, so convert back:
10 rad/s × 9.549297 r/min per rad/s = 95.49297 r/min.
Set the roller to 95.5 r/min. Two chances to go wrong here, and both are worth a deliberate check: using 0.250 m instead of the radius halves the answer, and stopping at step 2 leaves it in rad/s, which is not what a drive is dialled in.
From Rotation and Linear Speed
7. Wheel to road speed: A wheel 24 in in diameter turns at 6.0 rad/s. How fast is the vehicle travelling, in mi/h?
Show worked solution
r = 12 in, which is exactly 1 ft, or12 in × 0.0254 m/in = 0.3048 m.v = ωr = 6.0 rad/s × 0.3048 m = 1.8288 m/s.
Then convert to mi/h:
1.8288 m/s × = 4.0909… mi/h
or 4.09 mi/h. Both factors used here are exact by definition: 1 in = 0.0254 m and 1 mi/h = 0.44704 m/s.
From Rotation and Linear Speed
8. Wheel angular acceleration: A vehicle accelerates at 3.2 m/s2 on wheels 0.62 m in diameter. What is the angular acceleration of one wheel, in rad/s2?
Show worked solution
r = 0.31 m.- Rearrange
a = αrintoα = a/r = 3.2 m/s2 ÷ 0.31 m = 10.3226 rad/s2.
The answer is 10.3 rad/s². Check the unit as well as the number: dividing m/s2 by m leaves 1/s2, and because the radian is a length over a length, rad/s2 is that same thing named for what it describes. The result assumes the wheel rolls without slipping.
From Torque
9. Torque at a pulley: A belt applies 240 N of tangential force to a pulley 300 mm in diameter. What torque does that put on the shaft, in N·m?
Show worked solution
- The pulley is given by diameter, so halve it:
r = 150 mm = 0.150 m. τ = rF = 0.150 m × 240 N = 36 N·m.
The answer is 36 N·m — lower-case m, because N·M is not a valid SI torque unit; a prefix cannot stand alone. Using the 300 mm diameter instead of the radius gives 72 N·m, exactly twice too much.
From Torque
10. Reading a torque rating: A motor is rated 42 lbf·ft. What is that rating in N·m?
Show worked solution
42 lbf·ft × = 56.944… N·m
The rating is 56.9 N·m. Read the given unit carefully before reaching for the factor: this motor rating is a torque and converts to the newton metre. This site writes energy as ft·lbf, converting to joules, but word order is only a convention; the quantity and context determine which is meant. The numerical factor happens to be the same 1.355818 either way, so the arithmetic will not warn you if you have read the wrong quantity.
From Torque
11. Force from a torque: A winch drum is 400 mm in diameter and its drive supplies 95 N·m. What tangential force does that produce at the drum surface?
Show worked solution
r = 200 mm = 0.200 m.- Rearrange
τ = rFintoF = τ/r = 95 N·m ÷ 0.200 m = 475 N.
The drive produces 475 N at the drum surface. Notice what the radius is doing: wind more cable onto the drum, the effective radius grows, and the same torque pulls with less force. That single rearrangement is the whole story of a longer wrench, a bigger steering wheel, and a larger pulley.
From Gear Ratios
12. Single-stage reduction ratio: An 18-tooth driver turns a 54-tooth driven gear. Ignore losses (η = 1.00). Find i and enter X for X:1. Round to 2 decimal places.
Show worked solution
i = driven/driver = 54/18 = 3:1.
Enter 3.00 for 3.00:1.
From Gear Ratios
13. Single-stage output speed: An 18-tooth driver at ω_in = 1350 rpm turns a 54-tooth gear. Ignore losses (η = 1.00). Find the output speed magnitude. Round to 1 decimal place.
Show worked solution
i = 54/18 = 3; ω_out = 1350/3 = 450 rpm.
Enter 450.0 rpm.
From Gear Ratios
14. Single-stage ideal torque: An 18-tooth driver supplies 6.4 lbf·ft to a 54-tooth gear. Ignore losses (η = 1.00). Find the output torque magnitude at steady load. Round to 2 decimal places.
Show worked solution
i = 3; τ_out = 6.4 × 3 = 19.2 lbf·ft.
Enter 19.20 lbf·ft.
From Gear Ratios
15. Speed increase with a torque loss: A 48-tooth driver turns a 16-tooth output at steady load. Input torque is 15 lbf·ft. The single mesh is 94% efficient; there are no other losses. Find output torque. Round to 2 decimal places.
Show worked solution
i = 16/48 = 1/3, a speed increase. τ_out = 15 × (1/3) × 0.94 = 4.7 lbf·ft.
Enter 4.70 lbf·ft.
From Gear Ratios
16. Multiply compound ratios: A20 → B50; B shares a shaft with C24 → D60. Numbers are tooth counts. Ignore losses (η = 1.00). Find overall i and enter X for X:1. Round to 2 decimal places.
Show worked solution
B and C turn together. i = (50/20) × (60/24) = 2.5 × 2.5 = 6.25:1.
Enter 6.25 for 6.25:1.
From Gear Ratios
17. Speed on a compound shaft: A20 → B50; B is fixed to C24, which drives D60. Input ω_A = 1500 rpm. Each of the two stages is 97% efficient; no other losses. Find the speed magnitude of C. Round to 1 decimal place.
Show worked solution
ω_B = 1500 × 20/50 = 600 rpm. C shares B’s shaft, so ω_C has the same magnitude, 600 rpm. Losses do not change this speed.
Enter 600.0 rpm.
From Gear Ratios
18. Torque reaching the second stage: A20 → B50; B shares a shaft with C24 → D60. Input torque is 8 lbf·ft at steady rotation. Each stage is 97% efficient; no other losses or intermediate loads. Find the torque delivered by the first stage to the shared shaft. Round to 2 decimal places.
Show worked solution
Stage 1 gives τ_B = 8 × (50/20) × 0.97 = 19.4 lbf·ft. This is the torque available to drive C.
Enter 19.40 lbf·ft.
From Gear Ratios
19. Losses do not slow the compound output: A20 → B50; B shares a shaft with C24 → D60. Input ω_A = 1500 rpm. Each stage is 97% efficient; no other losses. Find output speed magnitude. Round to 1 decimal place.
Show worked solution
i = 2.5 × 2.5 = 6.25. ω_D = 1500/6.25 = 240 rpm. Do not multiply speed by 0.97².
Enter 240.0 rpm.
From Gear Ratios
20. Compound output torque with losses: A20 → B50; B shares a shaft with C24 → D60. Input torque is 8 lbf·ft at steady rotation. Each stage is 97% efficient; no other losses. Find output torque. Round to 3 decimal places.
Show worked solution
i = 6.25; η_total = 0.97² = 0.9409. τ_out = 8 × 6.25 × 0.9409 = 47.045 lbf·ft. Keep both efficiency factors.
Enter 47.045 lbf·ft.
From Gear Ratios
21. One idler: signed output speed: External gears A22 → I33 → B55 turn from an input of 1100 rpm clockwise. I is an idler. Ignore all losses, including both idler meshes (η = 1.00). Find signed output speed: clockwise positive, viewed from the same end of every shaft. Round to the nearest whole rpm.
Show worked solution
The idler count cancels: i = 55/22 = 2.5. Two meshes reverse twice, so output is clockwise: +440 rpm.
Enter 440 rpm.
From Gear Ratios
22. Two idlers: signed output speed: External gears A18 → I30 → J42 → B54 turn from an input of 900 rpm clockwise. I and J are idlers. Ignore all three mesh losses (η = 1.00). Find signed output speed, clockwise positive from the same end of every shaft. Round to the nearest whole rpm.
Show worked solution
i = 54/18 = 3. Three external meshes reverse direction three times: output is counterclockwise, so ω_out = −900/3 = −300 rpm.
Enter -300 rpm.
From Gear Ratios
23. Two stages, one idler, three losses: A20 → I30 → B60; B shares a shaft with C24 → D48. I is an idler. Every actual mesh is 95% efficient, including both meshes around I; there are no other losses. What is total efficiency as a percentage? Round to 2 decimal places.
Show worked solution
There are two reduction stages but three actual meshes. η_total = 0.95 × 0.95 × 0.95 = 0.857375 = 85.7375%.
Enter 85.74 %.
From Gear Ratios
24. An idler adds a specified mesh loss: A20 → I30 → B60; B shares a shaft with C24 → D48. I is an idler. Input torque is 7 lbf·ft at steady rotation. Every actual mesh is 95% efficient, including both meshes around I; no other losses. Find output torque. Round to 2 decimal places.
Show worked solution
i = (60/20) × (48/24) = 6. Three meshes give η_total = 0.95³. τ_out = 7 × 6 × 0.857375 = 36.00975 lbf·ft. The idler adds no reduction factor.
Enter 36.01 lbf·ft.
From Gear Ratios
25. Three stage efficiencies multiply: Three compound reduction stages each have one mesh and are each 96% efficient. There are no idlers or other losses. What is total efficiency as a percentage? Round to 2 decimal places.
Show worked solution
η_total = 0.96 × 0.96 × 0.96 = 0.884736 = 88.4736%. Neither 96% nor 288% is the total efficiency.
Enter 88.47 %.
From Gear Ratios
26. Total loss from three efficient stages: Three compound stages each have one mesh and are each 98% efficient. There are no idlers or other losses. What percentage of input power is lost overall? Round to 2 decimal places.
Show worked solution
η_total = 0.98³ = 0.941192. Total loss = (1 − 0.941192) × 100 = 5.8808%. Adding three 2% losses gives 6%, which is only an approximation.
Enter 5.88 %.
From Gear Ratios
27. Unequal efficiencies still multiply: Three stages have efficiencies 94%, 97% and 99%, respectively, with one mesh each. There are no idlers or other losses. Find total efficiency as a percentage. Round to 2 decimal places.
Show worked solution
η_total = 0.94 × 0.97 × 0.99 = 0.902682 = 90.2682%. Do not average the three percentages.
Enter 90.27 %.
From Gear Ratios
28. Three-stage torque versus ideal torque: A compound train has stage reductions 2:1, 3:1 and 2:1. Input torque is 5 lbf·ft at steady rotation. Each stage has one mesh at 96% efficiency; no idlers or other losses. Find output torque with losses. Round to 2 decimal places.
Show worked solution
i = 2 × 3 × 2 = 12. Ideal output torque is 60 lbf·ft; with losses, τ_out = 60 × 0.96³ = 53.08416 lbf·ft.
Enter 53.08 lbf·ft.
From Gear Ratios
29. Three-stage speed stays tied to teeth: A compound train has stage reductions 2:1, 3:1 and 2:1, with input ω_in = 1800 rpm. Each stage has one mesh at 96% efficiency; no idlers or other losses. Find output speed magnitude. Round to 1 decimal place.
Show worked solution
i = 12; ω_out = 1800/12 = 150 rpm. Efficiency changes the delivered torque and power, not the positive-drive speed ratio.
Enter 150.0 rpm.
From Gear Ratios
30. Power remaining after three stages: A gear train receives 1600 W. Three stages each have one mesh at 96% efficiency; no idlers or other losses. Find delivered output power. Round to 2 decimal places.
Show worked solution
P_out = 1600 × 0.96³ = 1415.5776 W. The stage ratios do not create power.
Enter 1415.58 W.
Keep at least 4 significant figures in η_total: 0.8847 gives 1415.52 W. The absolute tolerance is 0.057600000001 W: the larger of the printed-answer error (0.0024 W) and the 4-significant-figure answer error (0.0576 W), plus 0.000000000001 W to protect the numerical boundary.
From Gear Ratios
31. Work backward through the losses: A two-stage train has overall i = 10:1 and must deliver 80 lbf·ft at steady rotation. Each stage has one mesh at 95% efficiency; no idlers or other losses. What input torque is required? Round to 3 decimal places.
Show worked solution
Rearrange τ_out = τ_in × i × η_total: τ_in = 80/(10 × 0.95²) = 8.8642659… lbf·ft. Divide by efficiency when finding the required input.
Enter 8.864 lbf·ft.
From Gear Ratios
32. Minimum stages for a winch: A motor supplies 1920 rpm and 18 lbf·ft. A winch needs at most 40 rpm and at least 800 lbf·ft. Gears may have any integer count from 10–60, with unlimited copies; every reduction stage is at most 6:1 and 97% efficient. No idlers or other losses. Find the minimum feasible number of stages. Enter a whole number.
Show worked solution
Speed requires i ≥ 1920/40 = 48. Two stages reach at most 6² = 36, so they fail. Three stages can work: 12 → 48, 15 → 60, 16 → 50 give i = 50, 38.4 rpm and 821.4057 lbf·ft. Thus three stages are sufficient as well as necessary.
Enter 3 stages.
From Gear Ratios
33. Choose a whole-tooth final gear: A winch design uses three compound stages: 12 → 48, 15 → 60, then a 16-tooth driver → an unknown output gear. Choose the final driven tooth count for overall i = 50:1. Available teeth: integers 10–60, unlimited copies; each stage ≤6:1 and 97% efficient, with no idlers or other losses. Enter a whole number.
Show worked solution
The first two ratios multiply to 16. The last ratio must be 50/16 = 3.125, so driven teeth = 16 × 3.125 = 50. All teeth are in range and the last stage is below 6:1. Efficiency does not alter the tooth ratio.
Enter 50 teeth.
From Gear Ratios
34. Check the winch speed ceiling: A motor at 1920 rpm drives compound stages 12 → 48, 15 → 60 and 16 → 50 teeth. Each stage is 97% efficient; no idlers or other losses. Find output speed to check a 40 rpm ceiling. Round to 2 decimal places.
Show worked solution
i = 4 × 4 × 3.125 = 50. ω_out = 1920/50 = 38.4 rpm, which passes the 40 rpm ceiling. Do not apply efficiency to speed.
Enter 38.40 rpm.
From Gear Ratios
35. Check the winch torque floor: A motor supplies 18 lbf·ft at steady rotation to compound stages 12 → 48, 15 → 60 and 16 → 50 teeth. Each stage is 97% efficient; no idlers or other losses. Find output torque to check an 800 lbf·ft floor. Round to 2 decimal places.
Show worked solution
i = 50; η_total = 0.97³ = 0.912673. τ_out = 18 × 50 × 0.912673 = 821.4057 lbf·ft, which passes the 800 lbf·ft floor. Ideal torque alone (900 lbf·ft) is not the check.
Enter 821.41 lbf·ft.
Keep at least 4 significant figures in η_total: 0.9127 gives 821.43 lbf·ft. The absolute tolerance is 0.024300000001 lbf·ft: the larger of the printed-answer error (0.0043 lbf·ft) and the 4-significant-figure answer error (0.0243 lbf·ft), plus 0.000000000001 lbf·ft to protect the numerical boundary.
From Gear Ratios
36. Find the ratio required by torque: For a three-stage winch train, input torque is 18 lbf·ft and required output torque is at least 800 lbf·ft. Each stage has one mesh at 97% efficiency; no idlers or other losses. Find the minimum overall ratio allowed by torque alone; enter X for X:1. Keep at least 4 significant figures in η_total until the final answer. Round to 3 decimal places.
Show worked solution
i ≥ 800/(18 × 0.97³) = 48.6970081… . For a 1920 rpm input and 40 rpm ceiling, speed requires only i ≥ 48, so torque is the tighter requirement. Select actual tooth pairs at or above the unrounded bound.
Enter 48.697 for 48.697:1.
Keep at least 4 significant figures in η_total: 800/(18 × 0.9127) = 48.695567… rounds to 48.696. The absolute tolerance is 0.001008068: the larger of the printed-answer error (about 0.000008068) and the 4-significant-figure answer error (about 0.001008068), rounded upward at the numerical boundary. Full precision is preferred and is needed when checking actual tooth choices against the torque floor; 4 significant figures in η_total are accepted for this answer rounded to 3 decimal places.
From Gear Ratios
37. Choose the fastest feasible candidate: A winch needs ≤40 rpm and ≥800 lbf·ft from a 1920 rpm, 18 lbf·ft motor. All three-stage candidates use 10–60 teeth, ≤6:1 per stage and 97% per stage, with no idlers or other losses. A: 12 → 48, 15 → 60, 16 → 48 (48:1). B: 12 → 48, 15 → 60, 16 → 50 (50:1). C: 12 → 36, 10 → 60, 20 → 60 (54:1). Enter X for the X:1 ratio of the fastest candidate that meets both limits. Round to 2 decimal places.
Show worked solution
A gives 40 rpm but only 788.549472 lbf·ft: it fails torque. B gives 38.4 rpm and 821.4057 lbf·ft: both pass. C gives 35.5556… rpm and 887.118156 lbf·ft: both pass, but it is slower. Choose B, 50:1. This compares the three supplied candidates, not every possible train.
Enter 50.00 for 50.00:1.
From Saw Blade Selection
Teeth across a quarter inch: A uniform 24 TPI blade crosses 0.250 in of continuous material along the tooth line. Estimate teeth engaged. Give one decimal place.
Show worked solution
0.250 in × 24 teeth/in = 6.0 teeth. This is a spacing estimate; actual engagement changes as the blade moves.
From Saw Blade Selection
Teeth across a decimal thickness: A uniform 18 TPI blade crosses 0.250 in of continuous material along the tooth line. Estimate teeth engaged to one decimal place.
Show worked solution
0.250 × 18 = 4.5 teeth. Do not divide thickness by TPI; that does not count tooth spacings.
From Saw Blade Selection
Convert thickness before using TPI: A uniform 12 TPI blade crosses 6.35 mm of continuous material. Use 25.4 mm = 1 in. Estimate teeth engaged to one decimal place.
Show worked solution
6.35 ÷ 25.4 = 0.250 in. Then 0.250 × 12 = 3.0 teeth. Multiplying millimeters directly by teeth per inch mixes units.
From Saw Blade Selection
Minimum calculated pitch: For 0.125 in of continuous contact, calculate the minimum TPI for an estimated three teeth. Give one decimal place. This asks for the mathematical threshold, not a catalog selection.
Show worked solution
3 ÷ 0.125 = 24.0 TPI. Check: 0.125 × 24 = 3 teeth.
From Saw Blade Selection
Select from available pitches: For this simplified exercise, the candidate uniform pitches are 10, 14, and 18 TPI. A continuous contact length is 0.200 in. Which is the coarsest candidate meeting the three-tooth estimate? Enter the TPI. Actual selection also needs the maker's chart.
Show worked solution
3 ÷ 0.200 = 15 TPI minimum. The available choice is 18 TPI: 0.200 × 18 = 3.6 teeth. The tempting 14 TPI choice gives only 2.8.
From Saw Blade Selection
Thin-wall tube versus solid bar: A thin-wall tube and a solid bar have the same outside diameter. Which selection approach fits the tube? Enter 1 or 2.
1: Count the whole outside diameter as solid engagement.
2: Use a tube/structural chart with wall thickness and overall size; expect finer teeth than for the solid bar.
Show worked solution
2. The hollow center is not material. Engagement changes around the tube, so neither outside diameter alone nor a fixed two-wall sum describes the whole cut.
From Saw Blade Selection
Chip space in soft material: A blade specification calls for widely spaced teeth, generous gullets, and a zero-rake form for chip clearance in a softer material. Which named form fits? Enter 1, 2, or 3.
1: Fine regular.
2: Skip.
3: Positive-rake hook.
Show worked solution
2: skip. The generous chip space and zero-rake description identify the form. Final pitch still depends on section size.
From Saw Blade Selection
Identify a positive-rake form: A maker recommends an aggressive positive-rake tooth form for a supported production cut. Which form does that describe? Enter 1, 2, or 3.
1: Regular.
2: Skip.
3: Hook.
Show worked solution
3: hook. Positive rake describes the face angle. It does not establish pitch, set, or one universal rake angle.
From Saw Blade Selection
Fine-pitch set for thin sections: For a thin section, a fine-pitch blade specification calls for groups of teeth that gradually change sideways displacement. Which set matches? Enter 1, 2, or 3.
1: Left-right-straight raker.
2: Wavy.
3: Left-right alternate.
Show worked solution
2: wavy. Groups create the wave of set. Raker includes an unset tooth; alternate switches sides at each tooth. Set alone does not determine suitable TPI.
From Bench Work with Hand Tools
Hand hacksaw cutting direction: For a conventional push-cut hand hacksaw, which setup is correct? Enter 1 or 2.
1: Teeth away from the handle; pressure on the forward stroke, eased on return.
2: Teeth toward the handle; heavy pressure in both directions.
Show worked solution
1. The teeth face the cutting stroke. This describes a conventional hand hacksaw, not a universal power-hacksaw stroke direction.
From Bench Work with Hand Tools
Choose a file for draw filing: A flat face needs light finishing by draw filing after rough shaping. Which is the better starting choice? Enter 1 or 2.
1: A coarse double-cut file chosen for rapid stock removal.
2: A suitable fine single-cut file, kept clean and flat.
Show worked solution
2. A fine single-cut file suits controlled draw filing. Keep the file crosswise to the travel and avoid rocking across the face.
From Bench Work with Hand Tools
Choose a file for stock removal: An accessible flat steel edge needs substantial hand shaping before finishing. Which file is the better starting choice? Enter 1 or 2.
1: A suitable bastard-grade double-cut flat file.
2: A fine round file chosen for finishing a small round opening.
Show worked solution
1. Match both shape and removal rate. Bastard describes coarseness; double-cut describes the crossing tooth pattern.
From Bench Work with Hand Tools
Respond to chips lodged in a file: A file begins leaving deep scratches, and chips are lodged between its teeth. What should happen next? Enter 1 or 2.
1: Increase pressure so the file forces the chips through.
2: Clear the teeth with a file card, then inspect the work and resume controlled filing.
Show worked solution
2. This is pinning. Trapped chips can score the work; extra force does not restore a clean cutting surface.
From Lead Screws
1. Lead: A triple-start screw has pitch 1.5 mm. What is its lead in mm per revolution?
Show worked solution
L = pitch × starts = 1.5 × 3 = 4.5 mm/rev. The 1.5 mm spacing is pitch; it is not the travel per turn for this screw.
From Lead Screws
2. Actuator speed: A single-start ACME actuator has 0.20 in pitch and turns at an actual 900 rpm. Its given efficiency is 60%. Find linear speed in in/min during steady engaged travel; the nut cannot rotate.
Show worked solution
L = 0.20 × 1 = 0.20 in/rev. Then v = 900 × 0.20 = 180 in/min (3 in/s). Applying 60% to speed would give 108 in/min; efficiency affects force, not the engaged motion ratio.
From Lead Screws
3. Actuator force: A single-start ACME actuator with 0.20 in pitch receives 12 lb-in (lbf·in) of torque. Given overall efficiency 60%, find actual axial force during steady motion against a resisting load. Round to the nearest lbf.
Show worked solution
F = 0.60 × 2π × 12 / 0.20 = 226.1947 ≈ 226 lbf. The theoretical force is 376.9911 lbf. Omitting 2π would give only 36 lbf.
From Lead Screws
4. Handle torque: A 120 N hand force acts perpendicular to a vise handle, 180 mm from the screw axis. Find screw input torque in N·m.
Show worked solution
From Torque, T = rF = 0.180 × 120 = 21.6 N·m. The supplied 180 mm is already the axis-to-hand radius; do not halve it.
From Lead Screws
5. Clamping estimate: An illustrative single-start square-thread vise has 4 mm pitch and receives 21.6 N·m. Assume a given overall tightening efficiency of 30%, not a measured typical vise value. Estimate force while tightening, in kN, to two decimal places.
Show worked solution
L = 0.004 m/rev. F = 0.30 × 2π × 21.6 / 0.004 = 10,178.7602 N = 10.18 kN. Using π = 3.14 gives 10.1736 kN, or 10.17 kN rounded; both are accepted. This is not a rated vise capacity or a residual static-force prediction. Keeping 4 mm beside N·m without conversion makes the result 1,000 times too small.
From Lead Screws
6. Free jaw travel: Before contacting the work, a vise's single-start 4 mm pitch screw turns at 24 rpm. Find jaw speed in mm/s. The nut is restrained from rotation and backlash has been taken up.
Show worked solution
v = 24 × 4 = 96 mm/min. Divide by 60: 1.6 mm/s. The jaw cannot keep travelling at that speed through a rigid clamped workpiece.
From Lead Screws
7. Table feed: A CNC positioning slide uses a triple-start screw of 1.5 mm pitch, rotating at an actual 480 rpm. Given efficiency is 55%. Find steady feed in mm/s with the nut restrained from rotating and clearance taken up.
Show worked solution
L = 1.5 × 3 = 4.5 mm/rev. v = 480 × 4.5 / 60 = 36 mm/s. Pitch instead of lead gives 12 mm/s; applying efficiency gives 19.8 mm/s; leaving minutes unconverted gives 2,160 with the wrong unit.
From Lead Screws
8. Available thrust: A triple-start screw of 1.5 mm pitch receives 0.75 N·m at steady speed against a resisting load. Given assembly efficiency is 55%. Find actual axial force to the nearest N. This is available screw thrust, before allocating force to guide friction or other loads; it is not a cutting-capacity rating.
Show worked solution
L = 1.5 × 3 / 1,000 = 0.0045 m/rev. F = 0.55 × 2π × 0.75 / 0.0045 = 575.9587 ≈ 576 N. Using pitch instead of lead gives 1,727.876 N; omitting 2π gives 91.667 N.
From Lead Screws
9. Required torque: A two-start screw with 2.5 mm pitch must deliver 800 N at steady speed. Its given efficiency is 50%. Find the input torque in N·m, to two decimal places.
Show worked solution
L = 2.5 × 2 = 5 mm = 0.005 m. T = FL/(2πη) = 800 × 0.005/(2π × 0.50) = 1.27324 ≈ 1.27 N·m. Divide by efficiency when solving for the required input.
From Screw Threads
10. Inch coarse callout: What pitch in inches is specified by 1/4-20 UNC? This checked designation appears in NBS Circular 479; the source entry and link are in Screw Threads.
Show worked solution
20 is threads per inch: pitch = 1/20 = 0.050 in. The 1/4 is nominal diameter. UNC identifies the coarse series.
From Screw Threads
11. Inch fine callout: Find the pitch of 1/4-28 UNF in inches, to four decimal places. The NBS Circular 479 designation is checked and linked in Screw Threads.
Show worked solution
pitch = 1/28 = 0.035714… ≈ 0.0357 in. At the same nominal diameter, the fine series has a smaller pitch than the coarse series.
From Screw Threads
12. Metric callout: What pitch in mm does M6 × 1.0 specify? The NAAMS standard entry is quoted and linked in Screw Threads.
Show worked solution
The pitch is written directly: 1.0 mm. M6 gives the metric family and nominal 6 mm diameter. Do not invert the pitch as though it were a threads-per-inch count.
From Screw Threads
13. One-way thrust: Which form deliberately uses an asymmetric working flank to favor heavy axial thrust in one direction? Choose the form, not a rated capacity: 1 = V, 2 = square, 3 = ACME/trapezoidal, 4 = buttress.
Show worked solution
4 — buttress. Its working flank and sloping back are intentionally different. Reverse motion is possible, but heavy loading on the other flank does not keep the same strength and efficiency advantages.
From Screw Threads
14. Repeated travel: A slide needs a power-screw form suited to repeated travel in both directions, easier machining than square, and a split-nut arrangement to take up wear clearance. Which conventional form best matches? 1 = V, 2 = square, 3 = ACME/trapezoidal, 4 = buttress.
Show worked solution
3 — ACME/trapezoidal. This trades some square-thread efficiency for practical manufacturing and nut-adjustment advantages. Actual wear life, efficiency and load rating still require assembly data.
From Section Views
Identify a straight complete cut. Which section type uses the straight cut through the entire cylinder shown here?
Show explanation
Full. One straight plane passes through the whole part. Both walls are sectioned.
From Section Views
Identify a quarter removal. Which section type produces this combined exterior and interior view?
Show explanation
Half. Removing one quarter of a symmetric part produces a view with half exterior and half section.
From Section Views
Identify a jogged cut. Which section type follows the jogged path through all three holes?
Show explanation
Offset. The parallel cutting portions are offset to reach staggered holes. The jogs are not edges on the manufactured plate.
From Section Views
Identify a local cutaway. Which section type exposes this passage within an irregular boundary?
Show explanation
Broken-out. A broken-out section removes only a local area to a selected depth in the existing view.
From Section Views
Identify an overlaid slice. The rectangular slice is turned into the drawing plane and placed on the parent bar. What is it?
Show explanation
Revolved section. A revolved section places the rotated cross section on the parent view.
From Section Views
Identify a separate slice. The slice is shown clear of the parent bar and identified by its cut letters. What is it?
Show explanation
Removed section. A removed section is placed apart from the parent view and linked to its location by identification.
From Section Views
Identify a rotated radial cut. One radial cutting portion is rotated into the other before drawing the result. Which section type is used?
Show explanation
Aligned. An aligned section rotates an angled cutting portion into alignment; the top view retains the actual angular locations.
From Section Views
Choose the rib hatch convention. A section plane passes lengthwise through a thin reinforcing rib. Which treatment usually avoids making the rib look like a solid wall?
Show explanation
Leave the lengthwise rib unhatched. A rib cut along its length is conventionally left unhatched; a transverse cut through its thickness is hatched.
From Section Views
Distinguish shaft cut directions. Which statement is appropriate for a solid shaft in an assembly drawing?
Show explanation
A longitudinal shaft is commonly unsectioned; a transverse section can be hatched. The convention depends on cut direction and context. It does not make all shafts immune to sectioning.
From Section Views
Match a cutting plane to its result. Follow the X–X arrows. Which candidate matches the section through this plate?
Show explanation
Result A. Result A. The straight plane crosses the left hole. The right hole lies in the portion removed toward the viewer, so it does not appear in this section.
From Section Views
Interpret the offset jog. The cutting plane jogs across solid material between two holes. Should that jog appear as a new physical edge in the resulting section?
Show explanation
No, the jog changes the imaginary cut path. The offset cutting path is a drawing device, not a step machined into the part.
From Auxiliary and Detail Views
Choose a true-shape view. An inclined face appears as a sloping line in the front view. Which view shows the complete face in true shape?
Show explanation
An auxiliary view looking perpendicular to that face. The projection plane must be parallel to the face. Enlargement does not remove foreshortening.
From Auxiliary and Detail Views
Choose the projector direction. For a primary auxiliary construction, the fold line is parallel to the edge view of the inclined face. How do you draw the projectors?
Show explanation
Perpendicular to the fold line. Projectors are normal to the auxiliary projection plane as represented by its fold line; their direction follows the chosen face.
From Auxiliary and Detail Views
Read a partial auxiliary. A partial auxiliary shows only the sloping face and nearby identifying geometry. What has changed from a complete auxiliary?
Show explanation
The coverage is limited; the projection direction is unchanged. Partial describes which region is shown, not a new projection system.
From Auxiliary and Detail Views
Read a detail scale. A detail is marked 2:1. What does that tell you?
Show explanation
The drawn lengths are twice actual lengths at the intended print scale. A 2:1 view is an enlargement. The actual feature size remains the stated dimension; resized screens or prints are not measuring tools.
From Auxiliary and Detail Views
Read a broken view. A continuous rod is drawn with two break lines and an overall dimension of 360 mm. What is its specified overall length?
Show explanation
360 mm. The dimension gives the full length. The break saves sheet space and does not remove length from the manufactured rod.
From AC and DC
Sine-wave peak. A sine wave is 72 V RMS. What is its positive peak? Give one decimal place.
Show worked solution
72 × √2 = 101.8 V. RMS and peak are not equal.
From AC and DC
Sine-wave RMS. A sine wave peaks at ±210 V. What is its RMS voltage? Give one decimal place.
Show worked solution
210 / √2 = 148.5 V. Do not use peak-to-peak voltage.
From AC and DC
Line loss at the lower voltage. An ideal DC comparison transmits 6,000 W at the sending end at 300 V through a total line resistance of 0.8 Ω. What is the line heating loss? Give one decimal place.
Show worked solution
I = 6000 / 300 = 20 A; loss = 20² × 0.8 = 320.0 W. The line resistance includes the complete current path.
From AC and DC
Line loss at the higher voltage. Repeat the ideal DC comparison at 3,000 V, still transmitting 6,000 W at the sending end with total line resistance 0.8 Ω. What is the line heating loss? Give two decimal places.
Show worked solution
I = 6000 / 3000 = 2 A; loss = 2² × 0.8 = 3.20 W. Ten times the voltage produces one hundredth the loss: compare 320 W at 300 V.
From Series and Parallel Circuits
Series current. An ideal 18 V DC source drives 40 Ω and 80 Ω in series with both switches closed. What current passes through the 80 Ω resistor? Give three decimal places.
Show worked solution
R_eq = 40 + 80 = 120 Ω. The current through each resistor is 18 / 120 = 0.150 A; it does not divide.
From Series and Parallel Circuits
Series voltage drop across 40 ohms. An ideal 18 V DC source drives 40 Ω and 80 Ω in series. What is the drop across 40 Ω? Give two decimal places.
Show worked solution
I = 18 / 120 = 0.15 A. The drop is 0.15 × 40 = 6.00 V. The remaining 12 V drops across 80 Ω.
From Series and Parallel Circuits
Series voltage drop across 80 ohms. An ideal 18 V DC source drives 40 Ω and 80 Ω in series. What is the drop across 80 Ω? Give two decimal places.
Show worked solution
I = 0.15 A; drop = 0.15 × 80 = 12.00 V. The two drops add to 18 V.
From Series and Parallel Circuits
Parallel branch current. An ideal 18 V DC source supplies parallel 45 Ω and 75 Ω resistors. What current passes through 75 Ω? Give three decimal places.
Show worked solution
Each branch has 18 V: I = 18 / 75 = 0.240 A. The other branch draws 0.400 A.
From Series and Parallel Circuits
Parallel supply current. An ideal 18 V DC source supplies parallel 45 Ω and 75 Ω resistors. What is the total supply current? Give three decimal places.
Show worked solution
Add branch currents: 0.400 + 0.240 = 0.640 A.
From Series and Parallel Circuits
Parallel equivalent resistance. Find the equivalent resistance of 45 Ω and 75 Ω in parallel. Give two decimal places.
Show worked solution
R_eq = (45 × 75) / (45 + 75) = 28.125 Ω, or 28.13 Ω. It is less than either resistor.
From Series and Parallel Circuits
Which series resistor dissipates more?. An ideal 18 V DC source drives 40 Ω and 80 Ω in series. Identify the resistor dissipating more power, then enter that power in watts to two decimal places.
Show worked solution
The 80 Ω resistor dissipates more: 0.15² × 80 = 1.80 W. The 40 Ω resistor dissipates 0.90 W.
From Series and Parallel Circuits
Which parallel resistor dissipates more?. An ideal 18 V DC source supplies parallel 45 Ω and 75 Ω resistors. Identify the resistor dissipating more power, then enter that power in watts to two decimal places.
Show worked solution
The 45 Ω resistor dissipates more: 18² / 45 = 7.20 W. The 75 Ω resistor dissipates 4.32 W.
From Measuring Electric Circuits
Voltmeter connection and reading. An ideal 18 V DC source drives 56 Ω and 91 Ω in series. Where must an ideal voltmeter connect to read the voltage of the 56 Ω resistor? State the connection, then enter its reading to two decimal places.
Show worked solution
Connect across the 56 Ω resistor, in parallel. I = 18 / (56 + 91) = 18 / 147 ≈ 0.1224 A, so the reading is 18 × 56 / 147 ≈ 6.86 V. The numeric checker checks the reading; compare your connection with this explanation.
From Measuring Electric Circuits
Ammeter connection and reading. An ideal 18 V DC source supplies parallel 45 Ω and 75 Ω branches. Where must an ideal ammeter connect to read only the 75 Ω branch current? State the connection, then enter its reading to three decimal places.
Show worked solution
Insert it in series within the 75 Ω branch. It reads 18 / 75 = 0.240 A. In the supply lead it would read 0.640 A. The numeric checker checks the reading; compare your connection with this explanation.
From Measuring Electric Circuits
Ohmmeter reads the unpowered network. The source is disconnected and stored energy discharged from a parallel network of 68 Ω and 120 Ω. An ideal ohmmeter is connected across the two network terminals. What does it read? Give two decimal places.
Show worked solution
Both paths remain connected. The meter reads their equivalent resistance: (68 × 120) / (68 + 120) = 43.404255… Ω, or 43.40 Ω, not the resistance of either part alone.
From Measuring Electric Circuits
Current from a shunt reading. A voltmeter reads 18 mV across a 0.060 Ω shunt installed in series with a DC load. Find the load current to three decimal places.
Show worked solution
18 mV = 0.018 V. I = 0.018 / 0.060 = 0.300 A. The shunt carries the load current; the voltmeter connects across the shunt.
From Choosing a Woodworking Machine
Choose a machine for resawing
A straight board must be divided through its width into two thinner boards. Which machine is the best starting choice, assuming adequate capacity?
- Miter saw
- Band saw
- Thickness planer
Show answer and reasoning
Choice 2. A band saw resaws the thickness into separate boards. A planer removes thickness as chips; it does not recover a second board.
From Choosing a Woodworking Machine
Choose a machine for an enclosed opening
A thin panel needs a delicate enclosed curved opening, with no access cut from its outside edge. Which route fits?
- Band saw entering from the edge
- Miter saw
- Scroll saw with its blade threaded through a starter hole
Show answer and reasoning
Choice 3. The scroll-saw blade can enter the enclosed region through a starter hole. A continuous band cannot be threaded through an ordinary starter hole.
From Choosing a Woodworking Machine
Repeat an edge profile
Several flat blanks need the same curved outside outline copied from a template. Which machine and tooling best fit?
- Router with suitable template-guided tooling
- Drill press with twist bit
- Thickness planer
Show answer and reasoning
Choice 1. A template supplies the path for suitable router tooling. The bit removes the wood to that path.
From Choosing a Woodworking Machine
Choose a bit for a blind recess
A wood part needs a clean-sided blind circular recess with a largely flat bottom. A small center mark is acceptable. Which bit best fits?
- Hole saw
- Twist bit
- Forstner bit
Show answer and reasoning
Choice 3. A Forstner bit removes the recess interior and leaves a largely flat bottom. A hole saw leaves a core, and an ordinary twist bit leaves a conical bottom.
From Milling Rough Lumber Square
Order the rough-stock milling steps
Choose the final-sizing sequence that first establishes references, then controls thickness, width and length. Any preliminary breakdown has already left the blank oversized.
- Plane → rip → face-joint → crosscut → edge-joint
- Face-joint → edge-joint → plane → rip → crosscut
- Crosscut to final length → rip → plane → edge-joint → face-joint
Show answer and reasoning
Choice 2. The flat face supports thickness control; the straight square edge supports width control. Final end cuts follow the established face and edge geometry.
From Milling Rough Lumber Square
Recognize a planer reference
A bowed board goes through a thickness planer without a flattening fixture or a previously flattened face. What is the best prediction?
- It must emerge flat because thickness is uniform
- The planer automatically makes one edge square
- It may retain a bow after the rollers release it
Show answer and reasoning
Choice 3. Roller pressure can temporarily flatten the board. Uniform thickness does not prove flatness; establish a reference face first.
From Bending and Laminating Wood
Recognize bent lamination
A curved rail is made from thin strips with grain along their length, bonded together over a form. Which description fits?
- Bent lamination; springback can still occur
- Veneering; the face layer alone supplies all support
- Steam bending; adhesive removes the need for a form
Show answer and reasoning
Choice 1. The strips bend separately and the cured bonds retain the assembled curve. The bend may still open after release, so form compensation needs a trial.
From How Saws Cut Wood
Count strips with kerf
A board has exactly 12 in of usable width. Each 3 in strip is removed with its own cut from a remaining offcut. Kerf is 0.125 in per cut. Ignore other allowances. What is the maximum whole number of strips?
Show answer and reasoning
3 strips. floor(12 / (3 + 0.125)) = 3. Three strips and three kerfs use 9.375 in; four would need 12.5 in. Ignoring kerf gives 4, which does not fit.
From How Saws Cut Wood
Calculate the remaining offcut
From a 10 in usable width, remove three strips, each 2 in wide, using three cuts with a 0.125 in kerf. No other material is removed. What offcut width remains? Enter inches to three decimal places.
Show answer and reasoning
3.625 in. 10 − 3(2) − 3(0.125) = 3.625. The kerfs remove 0.375 in. Ignoring them would give 4 in.
From How Saws Cut Wood
Divide a board into equal final strips
An exactly 12 in wide board is divided into four equal strips with three cuts, each with a 0.125 in kerf. Both outside edges are already finished, no offcut remains, and no further allowance is needed. How wide is each strip? Enter inches to three decimal places.
Show answer and reasoning
2.906 in to three decimal places. (12 − 3 × 0.125) / 4 = 2.90625 in. Four pieces have three intervening cuts. Ignoring kerf gives 3 in; counting four cuts gives 2.875 in. Both are outside the answer tolerance.
From Fully Defined Sketches and Design Intent
Predict centered rectangle growth. An axis-aligned 64 × 22 mm rectangle has its center at the origin, and both sizes are driving. Increase width to 80 mm while preserving that intent. What is the new left-edge x coordinate in mm? Use a negative value for left of the origin.
Show answer and reasoning
−40 mm. Centering keeps the left and right coordinates opposite: −80/2 and +80/2. Each edge moves 8 mm outward from its original ±32 mm position. The new driving width gives a new fully defined rectangle; no free motion remains.
From Fully Defined Sketches and Design Intent
Measure a rectangle that can slide. Use the lab's axis-aligned rectangle model with four variables: left x, right x, bottom y and top y. Width and height drive size; bottom y is anchored at 0, but there is no horizontal anchor. Add a top-edge reference width measurement. How many DOF remain?
Show answer and reasoning
1 DOF. The three independent conditions fix width, height and vertical position. Horizontal translation remains. The reference dimension measures the width; it adds no constraint.
From Fully Defined Sketches and Design Intent
Distinguish repetition from contradiction. In the axis-aligned rectangle lab, a driving dimension sets bottom-edge width to 64 mm. Add a driving top-edge width of 74 mm without removing the first. What does this mean?
- One more independent condition makes the rectangle more fully defined
- The widths contradict; no rectangle satisfies both
- The second dimension is an agreeing redundant condition
Show answer and reasoning
2. Both edges measure right x − left x, so they cannot be 64 and 74 mm simultaneously. A second 64 mm condition would agree but add no independent rank. A reference width would only measure the result. This mathematical distinction does not prescribe a particular CAD application's error handling.
From Geometric Relations in Sketches
Two circles keep one center. The larger circle can change diameter without changing the smaller circle. Which single relation keeps their centers together?
- Equal
- Concentric
- Tangent
Show answer and reasoning
2 — Concentric. It shares the center while leaving the two radii independent. Equal would control size instead.
From Geometric Relations in Sketches
Separated edges on one line. Two nonzero straight edges must stay on the same extended line, with a gap allowed between their ends. What is this geometric relation called?
- Parallel
- Coincident endpoints
- Collinear
Show answer and reasoning
3 — Collinear. Parallel alone allows an offset between the lines; coincident endpoints would close the gap. In Onshape, point-on-line constraints can express collinearity.
From Geometric Relations in Sketches
A line touches a round boundary. A straight edge must keep the shown contact with the circle without crossing it. Which relation controls that contact?
- Tangent
- Equal
- Concentric
Show answer and reasoning
1 — Tangent. The radius to contact is perpendicular to the line. Equal does not control contact, and concentric applies to centers.
From Geometric Relations in Sketches
A locator point stays halfway. The point must remain halfway between the segment endpoints when the segment grows. Which relation captures that rule?
- Coincident with line
- Midpoint
- Fix
Show answer and reasoning
2 — Midpoint. A point merely coincident with the line may slide along it. Fix would hold its current location rather than follow the moving midpoint.
From Geometric Relations in Sketches
Separated arcs share a supporting circle. These two circular arcs must share both center and radius, but their endpoints stay separate. What is the SOLIDWORKS relation name?
- Concentric
- Equal
- Coradial
Show answer and reasoning
3 — Coradial. Concentric alone leaves radius free; equal alone leaves center position free. In Onshape, combine concentric and equal radius.
From Geometric Relations in Sketches
Two ribs change length together. Two separate straight segments represent layout ribs. Their lengths must stay equal, while their locations and directions may differ. Which relation should you use?
- Equal
- Parallel
- Collinear
Show answer and reasoning
1 — Equal. It links the lengths. Parallel links directions, and collinear puts the segments on the same supporting line.
From Fully Defined Sketches and Design Intent
An isolated sketch point. A single point lies in a 2D sketch. It has no relations, dimensions, or automatic constraints. How many degrees of freedom remain?
Show answer and reasoning
2 — Its x and y coordinates can vary independently.
From Fully Defined Sketches and Design Intent
A sized circle that can move. An isolated circle has one driving diameter of 16 mm. Its center has no constraints. How many DOF remain?
Show answer and reasoning
2 — The diameter removes the radius freedom from the original three. Center x and center y remain.
From Fully Defined Sketches and Design Intent
A circle centered at the origin. An isolated circle center is coincident with the origin. There is no radius or diameter dimension and no other constraint. How many DOF remain?
Show answer and reasoning
1 — The center is located in both directions. Only radius remains free.
From Fully Defined Sketches and Design Intent
A free horizontal segment. A nonzero line segment has only a horizontal relation. Neither endpoint is anchored and there is no length dimension. How many DOF remain?
Show answer and reasoning
3 — Starting x, starting y, and length remain. Equivalently, four endpoint coordinates minus one independent horizontal condition gives three.
From Fully Defined Sketches and Design Intent
A horizontal segment starts at the origin. A nonzero horizontal segment has its left endpoint coincident with the origin. No length dimension or other relation is present. How many DOF remain?
Show answer and reasoning
1 — The free endpoint can move horizontally. The origin coincidence removes two coordinate freedoms and horizontal removes one more from the original four.
From Fully Defined Sketches and Design Intent
A rectangle with location and width. A rectangle has joined corners and horizontal top/bottom and vertical sides. Its lower-left corner is coincident with the origin and a 52 mm driving dimension sets its width. Height is positive and undimensioned. There are no other constraints. How many DOF remain?
Show answer and reasoning
1 — Only height remains. The stated relations already control orientation, closure, and rectangular shape; the origin fixes location and the width dimension fixes size in x.
From Fully Defined Sketches and Design Intent
Finish locating a sized circle. A circle has a driving diameter and a vertical relation between its center and the origin. It should be centered 14 mm above the origin. Which single addition fully defines it?
- A radius dimension
- A 14 mm vertical center-to-origin dimension
- A horizontal relation between the center and origin
Show answer and reasoning
2 — A vertical center-to-origin distance of 14 mm removes the remaining vertical motion. Radius repeats size control. A horizontal alignment would put the center at the origin, against the stated intent.
From Fully Defined Sketches and Design Intent
Finish the length of an anchored edge. A horizontal line starts at the origin and extends right. It should be 37 mm long. Which addition removes the remaining freedom?
- A 37 mm driving length
- A second horizontal relation
- A vertical relation on the line
Show answer and reasoning
1 — Length is the only free variable. Horizontal is already present. Vertical would conflict with a nonzero horizontal line.
From Fully Defined Sketches and Design Intent
Match a second circle to a known size. Circle A is fully defined with a 12 mm diameter. Circle B has its center fully located elsewhere and must always match A in size. Which single relation fully defines B and captures that intent?
- Concentric
- Tangent
- Equal
Show answer and reasoning
3 — Equal removes B's radius freedom and keeps both circles the same size when A changes. Concentric conflicts with the intended separate centers. Tangent adds a different geometric requirement.
From Fully Defined Sketches and Design Intent
Finish a located label blank. An axis-aligned rectangle has joined corners, its lower-left corner at the origin, and a 68 mm driving width. It should be 26 mm tall. Which addition completes it?
- A second 68 mm driving dimension on the top edge
- A 26 mm driving height
- Fix the upper-right corner at its current position
Show answer and reasoning
2 — The height is the remaining independent size. The top edge is already 68 mm because of the rectangle's existing shape constraints. Fixing the upper-right corner would freeze an arbitrary current height rather than specify 26 mm.
From Fully Defined Sketches and Design Intent
Spot duplicate circle size control. A circle is fully defined by its center at the origin and a driving diameter of 28 mm. Which proposed driving dimension duplicates its existing size control?
- A 14 mm driving radius
- A driven diameter measurement
- A driven radius measurement
Show answer and reasoning
1 — The 14 mm radius is already determined by the 28 mm diameter. Keep it driven if you only need to report it. A driven diameter or radius is a measurement, not another driving constraint.
From Fully Defined Sketches and Design Intent
Spot a conflicting angle. A nonzero line has a horizontal relation. Which addition contradicts that relation?
- A driving length of 42 mm
- A driving vertical location of one endpoint
- A driving 25-degree angle to the sketch horizontal axis
Show answer and reasoning
3 — A 25-degree driving angle to the horizontal axis conflicts with horizontal. A length and a vertical position of an endpoint can coexist with horizontal, assuming no other constraints.
From Sketch Entities and Planes
Choose a sketch reference. You are starting a 2D profile for a new part with no existing faces. Which choice gives the sketch a plane and a useful position reference?
- Select only the origin and treat it as a plane
- Choose a standard plane and intentionally relate a sketch point to its origin
- Draw at an arbitrary screen location and assume that anchors the sketch
Show answer and reasoning
2 — A standard plane supplies orientation and placement in 3D; coincidence with its origin can anchor a point in that plane. The origin alone is not a plane, and screen appearance does not define a stable reference.
From Making and Inspecting Solder Joints
Identify poor wetting. Solder sits in a bead beside the lead and does not bond to it. Which diagnosis fits? Enter the choice number.
- A sound joint because solder is present
- A poorly wetted joint
- A properly insulated splice
Show explanation
2. A poorly wetted joint. The bead has not wetted both surfaces. Correct heat transfer and surface preparation before remaking the joint.
From Making and Inspecting Solder Joints
Judge a dull joint. A lead-free joint is dull but has a continuous wetted fillet and no bridge or crack. What follows from dullness alone? Enter the choice number.
- It must be cold
- It must be reheated until shiny
- Dullness alone does not prove a defect
Show explanation
3. Dullness alone does not prove a defect. Some lead-free alloys have a dull finish. Inspect wetting, shape, clearance, and damage.
From Making and Inspecting Solder Joints
Find an unintended connection. A ribbon of solder joins two pads that the circuit keeps separate. What is it? Enter the choice number.
- A solder bridge
- A Western Union splice
- A correct fillet
Show explanation
1. A solder bridge. A bridge makes an unintended electrical connection. Remove it before powering the circuit.
From Making and Inspecting Solder Joints
Identify movement during cooling. A lead shifted while the solder solidified, leaving a cracked joint. What is the defect? Enter the choice number.
- Too little insulation
- A disturbed joint
- Normal lead-free dullness
Show explanation
2. A disturbed joint. Movement during solidification can disrupt the joint. Secure the parts before reworking it.
From Making and Inspecting Solder Joints
Joint sequence: step 1. The bench and iron are ready. In the eight-step joint checklist, what is step 1? Enter the choice number.
- Heat both parts
- Strip and clean
- Remove solder, then iron
Show explanation
2. Strip and clean. Step 1 is: Strip and clean. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 2. You have completed step 1: strip and clean. In the eight-step joint checklist, what is step 2? Enter the choice number.
- Support the parts
- Hold still while cooling
- Feed solder into the heated joint
Show explanation
1. Support the parts. Step 2 is: Support the parts. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 3. You have completed step 2: support the parts. In the eight-step joint checklist, what is step 3? Enter the choice number.
- Inspect the joint
- Remove solder, then iron
- Flux if needed
Show explanation
3. Flux if needed. Step 3 is: Flux if needed. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 4. You have completed step 3: flux if needed. In the eight-step joint checklist, what is step 4? Enter the choice number.
- Hold still while cooling
- Heat both parts
- Strip and clean
Show explanation
2. Heat both parts. Step 4 is: Heat both parts. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 5. You have completed step 4: heat both parts. In the eight-step joint checklist, what is step 5? Enter the choice number.
- Feed solder into the heated joint
- Support the parts
- Inspect the joint
Show explanation
1. Feed solder into the heated joint. Step 5 is: Feed solder into the heated joint. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 6. You have completed step 5: feed solder into the heated joint. In the eight-step joint checklist, what is step 6? Enter the choice number.
- Flux if needed
- Strip and clean
- Remove solder, then iron
Show explanation
3. Remove solder, then iron. Step 6 is: Remove solder, then iron. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 7. You have completed step 6: remove solder, then iron. In the eight-step joint checklist, what is step 7? Enter the choice number.
- Support the parts
- Hold still while cooling
- Heat both parts
Show explanation
2. Hold still while cooling. Step 7 is: Hold still while cooling. Follow this order so preparation comes before heat, and cooling before inspection.
From Making and Inspecting Solder Joints
Joint sequence: step 8. You have completed step 7: hold still while cooling. In the eight-step joint checklist, what is step 8? Enter the choice number.
- Inspect the joint
- Feed solder into the heated joint
- Flux if needed
Show explanation
1. Inspect the joint. Step 8 is: Inspect the joint. Follow this order so preparation comes before heat, and cooling before inspection.
From Soldering Equipment and Setup
Meet a lead-free requirement. An assembly specification requires lead-free solder. Which spool fits that constraint? Enter the choice number.
- Tin-lead solder because it melts easily
- Any spool with rosin core
- The specified lead-free alloy with compatible electronics flux
Show explanation
3. The specified lead-free alloy with compatible electronics flux. Flux core does not identify the alloy. Match both the alloy requirement and the flux system.
From Soldering Equipment and Setup
Compare common alloy melting behavior. A repair specification permits tin-lead solder. Compared with common tin-copper or tin-silver-copper alloys, which generally melts at a lower temperature? Enter the choice number.
- Common tin-lead solder
- Every lead-free solder
- They always melt at the same temperature
Show explanation
1. Common tin-lead solder. The comparison is between these common alloy families, not every alloy sold as lead-free. Follow the repair specification.
From Soldering Wire Splices and Through-Hole PCBs
Name an inline wrapped splice. Each free wire end wraps around the opposite standing wire, making an inline connection. Name it. Enter the choice number.
- Lap splice
- Twist pigtail
- Western Union splice
Show explanation
3. Western Union splice. Western Union uses opposing wraps. A lap overlaps straight ends; a pigtail twists ends together on the same side.
From Soldering Wire Splices and Through-Hole PCBs
Name an overlapping splice. Two stripped ends overlap side by side without wrapping. Name the arrangement. Enter the choice number.
- Lap splice
- Western Union splice
- Twist pigtail
Show explanation
1. Lap splice. This is a lap or lay-over splice. It needs support during soldering and strain relief in use.
From Soldering Wire Splices and Through-Hole PCBs
Name a pigtail splice. Two stripped ends twist together into a joined tail on the same side. Name it. Enter the choice number.
- Lap splice
- Twist or pigtail splice
- Western Union splice
Show explanation
2. Twist or pigtail splice. The joined tail distinguishes this arrangement from an inline Western Union splice.
From Soldering Equipment and Setup
Apply the battery shop rule. A connection would require putting a soldering iron directly on a battery. What should you do? Enter the choice number.
- Work faster
- Use more flux
- Do not solder the battery; use its intended holder or connector
Show explanation
3. Do not solder the battery; use its intended holder or connector. NEVER SOLDER BATTERIES. Heat can damage a cell; the rule is not conditional on speed or flux.
From Soldering Equipment and Setup
Keep flux fume out of the breathing zone. Flux fume rises toward your face. What action fits the setup guidance? Enter the choice number.
- Use effective extraction and keep your head out of the plume
- Switch to lead-free and ignore the fume
- Blow it toward the next person
Show explanation
1. Use effective extraction and keep your head out of the plume. Lead-free solder still uses flux. Arrange ventilation or extraction so fume is kept away from people.
From Soldering Equipment and Setup
Handle leaded solder cleanly. You have handled leaded solder and are leaving for a snack. Choose the correct action. Enter the choice number.
- Eat at the soldering bench
- Wash your hands and eat away from the bench
- Wipe your hands on your clothing
Show explanation
2. Wash your hands and eat away from the bench. Wash after handling leaded solder, and keep food and drink away from the bench.
From Soldering Wire Splices and Through-Hole PCBs
Heat a through-hole connection. Where should the iron make contact for a through-hole PCB joint? Enter the choice number.
- Only the component lead
- Only the board surface beside the pad
- Both the copper pad and component lead
Show explanation
3. Both the copper pad and component lead. Both surfaces must get hot enough for solder to wet them. Use a tinned working face that fits the joint.
From Soldering Wire Splices and Through-Hole PCBs
Prepare insulation before soldering. Both ends of an inline cable will be inaccessible after the splice is closed. When should heat-shrink be slipped onto a wire? Enter the choice number.
- Before joining, kept away from heat until the joint cools and is inspected
- While solder is molten
- Only after both ends become inaccessible
Show explanation
1. Before joining, kept away from heat until the joint cools and is inspected. Put the tubing on first, away from soldering heat. Cover and shrink it after cooling and inspection.
From Material Utilization and Cost
Count after excluded trim. A 1,000 mm bar has 10 mm left and 20 mm right excluded trim, each already including its own cut loss. Blanks are 100 mm long. Reserve one full 2 mm separating kerf after every blank, including the last. All blanks are acceptable and have equal cross-section. How many whole blanks fit under this policy?
Show worked solution
Usable interval = 1,000 − 10 − 20 = 970 mm. Each blank plus its kerf uses 102 mm. floor(970 / 102) = 9 blanks. Ten need 1,020 usable mm. Do not add trim cut losses again.
From Material Utilization and Cost
Close the stock length balance. A 1,000 mm bar has 30 mm total excluded trim, already including trim cut losses. Nine acceptable 100 mm blanks each require a full 2 mm separating kerf, including the last blank. How long is the remaining offcut?
Show worked solution
Offcut = 1,000 − 30 − 9 × 100 − 9 × 2 = 52 mm. Check: 30 + 900 + 18 + 52 = 1,000 mm. Counting only eight kerfs would incorrectly give 54 mm.
From Material Utilization and Cost
Find the next blank threshold. Blanks are 98 mm long. Reserve one full 3 mm separating kerf after every blank, including the last. With no excluded trim, what minimum stock length supplies ten whole blanks under this policy?
Show worked solution
Ten blank-and-kerf intervals need 10 × (98 + 3) = 1,010 mm. At 1,009.99 mm only nine fit. The n − 1 kerf method gives 1,007 mm, which belongs to a different final-cut policy.
From Material Utilization and Cost
Allocate stock cost after a count drop. A $48 stock bar yields nine identical acceptable blanks under the stated layout. Charge the whole stock price equally to these blanks, with no remnant credit or other charges. Find stock-only cost per acceptable blank to the nearest cent.
Show worked solution
$48 / 9 = $5.333… = $5.33 per blank. Continuing to divide by ten would give $4.80 and leave part of the stock charge unassigned.
From Material Utilization and Cost
Use purchased length in utilization. Nine acceptable 100 mm blanks are made from a 1,000 mm bar of the same cross-section. Excluded trims, separating kerfs and offcut are outside the blanks. What is utilization to one decimal place, based on all purchased material?
Show worked solution
Acceptable blank length = 9 × 100 = 900 mm. Utilization = 900 / 1,000 × 100 = 90.0%. Dividing by only a 970 mm prepared interval would give 92.8%, a different denominator.
From Material Utilization and Cost
Interpret a zero-blank layout. A $12, 100 mm bar must supply 98 mm blanks with a full 3 mm separating kerf after every blank, including the last. There is no excluded trim. Which statement follows under this policy? Enter the choice number.
- One blank fits and costs $12.
- No blank fits, and stock cost per acceptable blank is undefined.
- No blank fits, so each acceptable blank costs $0.
Show worked solution
Choice 2. One blank-and-kerf interval needs 101 mm, so floor(100 / 101) = 0. The stock still costs $12, but division by zero blanks does not define a per-blank cost.
From Value Stream Mapping
First completion with transfer-all. An order has 10 jobs. A takes 4 min/job, B takes 6. All ten must finish A before the batch transfers. When does the first job finish B?
Assume all equal jobs arrive at time 0, A and B each handle one job at a time in order, A never blocks and waiting space is unlimited. Transfer is immediate when a batch completes; setup, transport, failure, rework and variation are zero.
Show worked solution
A finishes ten jobs at 10 × 4 = 40 min. B then needs 6 min for the first job: 46 min.
From Value Stream Mapping
Full order with transfer-one. Ten jobs take 4 min each at A and 6 min each at B. Each completed job transfers immediately. When does the whole order finish B?
Assume all equal jobs arrive at time 0, A and B each handle one job at a time in order, A never blocks and waiting space is unlimited. Transfer is immediate when a batch completes; setup, transport, failure, rework and variation are zero.
Show worked solution
The first job reaches B at 4 min. A supplies the later jobs faster than B uses them, so B runs without a gap for 10 × 6 min: 4 + 60 = 64 min.
From Value Stream Mapping
A final partial transfer batch. Five jobs take 3 min each at A and 2 min each at B. Transfer in batches of two; the final single job transfers as soon as it finishes A. When does the whole order finish B?
Assume all equal jobs arrive at time 0, A and B each handle one job at a time in order, A never blocks and waiting space is unlimited. Transfer is immediate when a batch completes; setup, transport, failure, rework and variation are zero.
Show worked solution
B finishes jobs 1–5 at 8, 10, 14, 16 and 18 min. The last job transfers at 15 min but B is occupied until 16 min.
From Value Stream Mapping
Separate processing from waiting. Ten jobs take 4 min at A and 6 min at B; transfer one at a time. For job 10, how much total waiting accumulates from order release 0 until completion?
Assume all equal jobs arrive at time 0, A and B each handle one job at a time in order, A never blocks and waiting space is unlimited. Transfer is immediate when a batch completes; setup, transport, failure, rework and variation are zero.
Show worked solution
Job 10 waits 36 min before A, 0 held for its batch and 18 at B: 54 min. Adding 10 min of processing gives its 64 min completion time.
From Value Stream Mapping
Processing time stays separate. Five jobs take 3 min/job at A and 2 min/job at B. Transfer in batches of two. How much active processing does each job need across A and B?
Assume all equal jobs arrive at time 0, A and B each handle one job at a time in order, A never blocks and waiting space is unlimited. Transfer is immediate when a batch completes; setup, transport, failure, rework and variation are zero.
Show worked solution
Processing is 3 + 2 = 5 min per job. Batch holding and station queues affect elapsed time, not these stipulated processing durations.
From Dimensional Analysis
1. Assembly-line travel time: Two stations are 10 m apart. Parts enter a conveyor 1 ft apart, and the line processes 2 parts every 15 s at a steady rate. How long does one part take to travel from the first station to the second?
Show worked solution
The spacing supplies 1 part / 1 ft, and the throughput supplies 15 s / 2 parts. Therefore
10 m × × × = 246.063 s
or 246 s (about 4.10 min). This assumes constant belt speed and steady spacing.
From Dimensional Analysis
2. Recycling-line loading: A recycling conveyor carries 72 lb of material per linear foot. It moves continuously at 0.23 m/s. How many lb of material pass a point in one hour?
Show worked solution
Carry the line loading, speed, exact foot factor, and hour conversion together:
0.23 m/s × × × = 195,590.551… lb/h
The result is 1.96 × 105 lb/h (about 195,600 lb/h).
From Dimensional Analysis
3. Typing count rate: A typist produces 150 words/min. The average word has 5.1 letters, and one spacebar press follows each word, for 6.1 key presses/word. What is the typist’s key rate in keys/s?
Show worked solution
150 words/min × × = 15.25 keys/s
Words and minutes cancel, leaving 15.25 keys/s.
From Motion Quantities
4. Average running speed: A runner completes 5 km in 12 min 37 s. What is the runner’s average speed in mi/h?
Show worked solution
First convert the distance and time:
5 km × = 3.10685596 mi
12 min 37 s = 757 s
Then
3.10685596 mi / 757 s × = 14.775… mi/h
or 14.8 mph.
From Dimensional Analysis
5. Conveyor setting: A packaging conveyor must cover 20 m in 12 s. What speed setting is required in ft/s?
Show worked solution
20 m / 12 s × = 5.468066… ft/s
The required speed is 5.47 ft/s.
From Dimensional Analysis
6. Reverse exact-factor check: A guard is specified as 18 ft long. What is its length in meters?
Show worked solution
Put feet on the bottom so they cancel:
18 ft × = 5.4864 m
The exact relationship gives 5.4864 m.
From Measurement Language
7. Prefix scale: A feature is 0.0048 m long. What is its length in mm?
Show worked solution
0.0048 m × = 4.8 mm
The answer is 4.8 mm.
From Geometry, Mass, and Force
8. Squared conversion: A panel has an area of 2.75 ft2. What is its area in in2?
Show worked solution
Because 1 ft = 12 in, square the relationship:
2.75 ft2 × = 2.75 × 144 in2 = 396 in2
The answer is 396 in².
From Geometry, Mass, and Force
9. Cubed conversion: A fixture has a volume of 0.50 ft3. What is its volume in in3?
Show worked solution
Cube the 12-in-per-foot relationship:
0.50 ft3 × = 0.50 × 1,728 in3 = 864 in3
The answer is 864 in³.
From Acceleration and Squared Time
10. Conveyor acceleration: A conveyor accelerates from rest (0 m/s) to 3 m/s in 5 s. What is its average acceleration?
Show worked solution
aavg = (vf − vi) / Δt = (3 m/s − 0 m/s) / 5 s = 0.6 m/s2. The answer is +0.6 m/s² in the selected positive direction.
From Acceleration and Squared Time
11. Direction reversal: Take motion to the right as positive. A cart changes from +2 m/s to −1 m/s in 0.75 s. What is its average acceleration?
Show worked solution
aavg = (−1 m/s − (+2 m/s)) / 0.75 s = −3 m/s / 0.75 s = −4 m/s2. The negative sign means the acceleration component points left, opposite the chosen positive direction.
From Dimensional Analysis
12. Complete-part schedule: A cell runs for 7.5 h. Setup takes 20 min, and each repeatable cycle takes 3.6 min per part. How many complete parts fit in the run?
Show worked solution
The running window is
7.5 h × − 20 min = 430 min
Then
430 min × = 119.444… parts
Only 119 complete parts fit; the fraction is time into the next cycle.
From Measurement Language
Practice: A feature is specified as 3,200 µm. What is that length in millimeters?
Show worked solution
Micro means 10−6 and milli means 10−3. Moving from micrometers to millimeters divides by 1,000:
3,200 µm × = 3.2 mm
From Dimensional Analysis
Practice: A SolidWorks model reports 400 in3. What is the volume in ft3?
Show worked solution
There are 12 in in 1 ft, so there are 123 = 1,728 in3 in 1 ft3. Therefore
400 in3 × = 0.23148148 ft3
Rounded sensibly: 0.2315 ft3.
From Motion Quantities
Practice: A cart moves at a constant 4.5 ft/s between two areas 0.25 mi apart. Ignoring loading and turnaround time, what is its idealized round-trip rate per hour?
Show worked solution
One round trip is 2 × 0.25 mi = 0.50 mi.
0.50 mi × = 2,640 ft
At 4.5 ft/s, one trip takes
2,640 ft / = 586.667 s
Therefore
3,600 s / 586.667 s = 6.13636
round trips per hour, or about 6.14 round trips per hour. A plan that requires complete trips would schedule 6 and leave time for only part of a seventh.
From Acceleration and Squared Time
Practice: A machine arm starts from rest and reaches 3 m/s in 4 s. What is its average acceleration, assuming the stated interval?
Show worked solution
aavg = (vf − vi) / Δt = (3 m/s − 0 m/s) / 4 s = 0.75 m/s2. The assumption is that the requested value is the average over the complete four-second interval.
From Geometry, Mass, and Force
Practice: A sheet-metal enclosure has 1,250 in2 of external surface area. A U.S. gallon of paint covers 350 ft2. Ignoring overspray, waste, unpainted areas, and extra coats, how many complete enclosures can one gallon cover?
Show worked solution
Convert the enclosure area using the squared relationship:
1,250 in2 × = 8.68056 ft2
Then
350 ft2 / 8.68056 ft2 per enclosure = 40.32 enclosures
Only 40 complete enclosures can be painted. The answer assumes the stated coverage is for the required coat and all paint reaches the enclosure.
From Wood Joinery and Fasteners
A supported shelf: A fixed shelf must be located across the grain of a solid side panel and have a shoulder beneath it. Which joint directly provides that seat?
- A plain end-to-face butt joint without reinforcement
- A fitted dado with enough side-panel thickness remaining
- A miter cut only on the shelf end
Show explanation
Choice 2. A dado provides a bearing surface under the shelf and shoulders for location. It still needs suitable proportions and a way to resist withdrawal.
From Wood Joinery and Fasteners
A moving tabletop: A wide solid-wood top crosses a narrow support rail. Which attachment can allow seasonal movement across the top’s grain?
- Glue the full width of the top to the rail
- Install rigid screws across the width with no travel
- Use suitable tabletop clips with the required movement clearance
Show explanation
Choice 3. Clips can hold the top while allowing travel. The movement direction and clearance must match the design; a finish does not remove that need.
A small gap in a repair: For a noncritical wood repair with a small gap that cannot be refitted, which candidate should be checked first for a structural gap-filling specification?
- Ordinary PVA with sawdust added
- An epoxy system with its specified bonding filler
- Foaming polyurethane chosen only because it expands
Show explanation
Choice 2. Investigate the epoxy system and its gap limit, substrate preparation and cure instructions. A structural bonding filler differs from an easily sanded fairing filler. Foam expansion alone is not evidence of strength.
Adhesive for damp service: A close-fitting wood joint will experience occasional wetting. What establishes whether the adhesive is suitable?
- The exact product’s exposure rating and application limits
- Its yellow color
- The fact that all wood glues are waterproof
Show explanation
Choice 1. Use the product’s stated service conditions, including limits on immersion, temperature and structural use. Adhesive family and color are not exposure ratings.
Unclamping versus loading: A product’s minimum clamp interval has elapsed, but its stated interval before loading has not. What follows from those instructions?
- The joint is ready for its full service load
- The glue must have failed
- Removing restraint and applying service loads are different milestones
Show explanation
Choice 3. Clamp time and cure/loading time are separate. Follow both instructions under the specified conditions rather than assuming that an unclamped joint has full cured properties.
From Preparing and Finishing Wood
Order the preparation grits: A sample calls for all four grits: 80, 120, 150 and 180. Which sequence progressively removes coarser scratches?
- 180 → 150 → 120 → 80
- 80 → 120 → 150 → 180
- 80 → 180 → 120 → 150
Show explanation
Choice 2. Use 80 → 120 → 150 → 180, inspecting for removal of the previous scratches at each step. This exercise specifies the grits; a real finishing system may call for a different start or endpoint.
From Preparing and Finishing Wood
Finish for frequent wiping: A surface will receive frequent wiping and light abrasion. The owner wants visible grain and accepts a film appearance. Which is the best starting candidate?
- A clear film finish explicitly rated for that service
- Pigment stain alone
- Wax alone chosen for maximum wear resistance
Show explanation
Choice 1. A suitable clear film can provide wear and spill protection while leaving grain visible. Check the exact formulation and complete preparation/cure schedule. Stain primarily supplies color; wax alone provides limited protection.
From Preparing and Finishing Wood
A level open-pore surface: A sample has open pores but no dents. A level high-gloss surface is wanted. Which preparation product directly addresses those pores?
- A screw-hole plug
- Pre-stain conditioner alone
- A compatible grain/pore filler
Show explanation
Choice 3. Pore filler levels the open texture. Conditioner addresses uneven stain absorption; it does not perform the same job. Test filler order and color in the complete finishing system.
From Drying and Conditioning Wood
Moisture from sample masses: A sample initially weighs 144 g. Its oven-dry mass is 120 g. Calculate MC on the oven-dry basis, to one decimal place.
Show explanation
Water mass = 144 − 120 = 24 g. MC = 24 / 120 × 100 = 20.0%. Dividing by the initial 144 g gives 16.7%, which is a different basis.
From Drying and Conditioning Wood
Round a moisture result: A sample initially weighs 125 g. Its oven-dry mass is 110 g. Calculate MC on the oven-dry basis, to one decimal place.
Show explanation
Water mass = 125 − 110 = 15 g. MC = 15 / 110 × 100 = 13.636…%, or 13.6%. Dividing by the initial 125 g gives 12.0%, which is incorrect for the requested basis.
From Buying Lumber and Board Feet
Explicit sales tally: A seller tallies a board using nominal thickness 1 in, measured width 7 in, and length 9 ft. Use exactly those dimensions. How many board feet? Give two decimals.
Show worked solution
1 × 7 × 9 / 12 = 5.25 BF. This is the stated tally basis, not a measurement of the finished volume.
From Buying Lumber and Board Feet
Finished physical volume: A finished board actually measures 3/4 in thick, 7 in wide, and 9 ft long. Express its physical volume in board feet to two decimals.
Show worked solution
0.75 × 7 × 9 / 12 = 3.9375 BF, or 3.94 BF. Substituting nominal 1 in thickness would answer a different question.
From Buying Lumber and Board Feet
Three boards: Three boards each actually measure 1.25 in × 5.5 in × 8 ft. Find their combined physical volume in board feet to two decimals.
Show worked solution
Each board is 1.25 × 5.5 × 8 / 12 = 4.583333… BF. Multiply the unrounded value by 3: 13.75 BF.
From Buying Lumber and Board Feet
Material allowance: Your parts need 30 BF of physical volume. You expect 20% of the material you buy to be lost to saw kerf, defects, flattening and layout. Using that convention, how many board feet of physical volume should you buy? Give two decimals.
Show worked solution
The 20% is a share of what you buy, so only 80% of the purchase becomes parts: 30 ÷ (1 − 0.20) = 30 ÷ 0.80 = 37.50 BF. Adding 20% to the need (30 × 1.20 = 36 BF) uses the other convention and leaves you short under this one.
From Buying Lumber and Board Feet
All dimensions in inches: A piece actually measures 2 in × 9 in × 40 in. Find its physical volume in board feet to two decimals.
Show worked solution
All dimensions are inches, so divide cubic inches by 144: 2 × 9 × 40 / 144 = 5.00 BF.
From Buying Lumber and Board Feet
Nominal and actual width: A dry surfaced nominal 2 × 4 is specified to have an actual cross-section of 1-1/2 × 3-1/2 in. What width belongs in a physical-volume calculation? Give two decimals.
Show worked solution
Use the supplied actual width, 3.50 in. The nominal width 4 in is its sales name.
From Wood Structure and Species
Classify balsa: Balsa dents easily. Is it botanically a hardwood or softwood? Enter 1 for hardwood or 2 for softwood.
Show worked solution
1 — hardwood. Balsa comes from an angiosperm tree. Hardness is not the classification test.
From Wood Structure and Species
Classify yellow pine: Southern yellow pine can be harder than balsa. Enter 1 if this makes it a hardwood, or 2 if it remains a softwood.
Show worked solution
2 — softwood. Pine is a gymnosperm; relative hardness does not change its botanical group.
From Grain and Moisture Movement
Identify cup: A board is trough-shaped across its width. Identify the defect: 1 = cup, 2 = bow, 3 = crook, 4 = twist, 5 = check.
Show worked solution
1 — Cup. Cup is curvature across the width.
From Grain and Moisture Movement
Identify bow: The face curves along its length, while both long edges bend together out of the face plane. Identify the defect: 1 = cup, 2 = bow, 3 = crook, 4 = twist, 5 = check.
Show worked solution
2 — Bow. Bow is lengthwise curvature out of the broad face plane.
From Grain and Moisture Movement
Identify crook: Viewed from above its broad face, a board bends sideways along its length. Identify the defect: 1 = cup, 2 = bow, 3 = crook, 4 = twist, 5 = check.
Show worked solution
3 — Crook. Crook is edgewise curvature within the face plane.
From Grain and Moisture Movement
Identify twist: The two ends rotate relative to each other so the four corners do not lie in one plane. Identify the defect: 1 = cup, 2 = bow, 3 = crook, 4 = twist, 5 = check.
Show worked solution
4 — Twist. Twist changes the relative orientation of the ends.
From Grain and Moisture Movement
Identify check: A narrow separation follows the grain on a face but does not pass through the board. Identify the defect: 1 = cup, 2 = bow, 3 = crook, 4 = twist, 5 = check.
Show worked solution
5 — Check. A check is a separation that does not extend through the piece; a through separation is a split.
Painted interior panel: Choose a smooth panel for a nonstructural painted interior decorative door with a routed profile, in dry service. 1 = ordinary MDF, 2 = rough structural OSB. Enter the better starting choice.
Show worked solution
1 — MDF. Fine fibers provide a smooth surface and consistent profiles. Plan suitable fastening; this does not make ordinary MDF a wet-service product.
Rated wall sheathing: A wall design calls for rated structural sheathing. Which is the relevant starting product? 1 = unrated furniture particleboard, 2 = OSB with the specified structural rating and thickness.
Show worked solution
2 — the specified rated OSB. The product rating, orientation, and installation details matter; a generic sheet name is not a design specification.
Laminate substrate: A dry interior furniture panel needs an economical substrate under decorative laminate, and its design already accommodates the substrate strength and fasteners. 1 = particleboard, 2 = glulam beam. Which is the suitable product form?
Show worked solution
1 — particleboard. It supplies a panel substrate. Glulam supplies structural members, not an interchangeable sheet surface.
From Grain and Moisture Movement
Center-fixed edge travel. A flat-sawn white fir board is 9.15 in wide when installed at 8% MC and reaches 11% MC. Use CT = 0.00245 per percentage point MC. Assume uniform free width change on a stationary support, with the center as the only fixed reference. How far does the right edge move outward? Give inches to four decimal places.
Show answer and reasoning
Total width change = 9.15 × 0.00245 × (11 − 8) = 0.0672525 in. The right edge is half a width from the reference, so it moves 0.0672525 ÷ 2 = 0.0336 in outward. The total width change is shared by two edges.
From Grain and Moisture Movement
Left-edge-fixed travel. A flat-sawn white fir board is installed 9.15 in wide at 8% MC. It reaches 11% MC; CT = 0.00245 per percentage point MC. Assume uniform free width change and a stationary support with the left edge as the only fixed reference. How far does the right edge move outward? Give inches to four decimal places.
Show answer and reasoning
The reference is one full width from the right edge. Travel = 9.15 × 0.00245 × 3 = 0.0673 in outward. Dividing by two would incorrectly use a center reference.
From Grain and Moisture Movement
Drying reverses travel. A flat-sawn white fir board is installed 9.15 in wide at 8% MC, then dries to 6% MC. Use CT = 0.00245 and assume uniform free width change with only the center fixed to a stationary support. Find signed right-edge travel: positive means outward, negative means inward toward the fixed reference. Give inches to four decimal places, including the sign.
Show answer and reasoning
ΔW = 9.15 × 0.00245 × (6 − 8) = −0.044835 in. Right-edge travel = ΔW ÷ 2 = −0.0224 in. This uses the inward allowance, even if ample outward allowance exists.
From Grain and Moisture Movement
Check the correct travel direction. An ideal center-fixed board's right edge needs 0.0224 in inward travel from its installed position. The stationary support allows 0.020 in inward and 0.040 in outward clear travel at that point. Which conclusion follows from this geometric comparison?
Show answer and reasoning
2. Compare 0.0224 with 0.020, not with their sum or the outward allowance. The excess is about 0.0024 in. The model estimates free displacement; it does not calculate stress, cracking or joint strength.
From Grain and Moisture Movement
A point inside the edge. A 9.15 in wide board has a calculated free width increase of 0.0672525 in. Assume uniform strain and a stationary support with only the center fixed. A tracking point is one quarter of the installed full width to the right of the center. How far does this point move outward? Give inches to four decimal places.
Show answer and reasoning
Point travel = (distance from reference ÷ installed width) × ΔW = 0.25 × 0.0672525 = 0.0168 in outward. The edge is farther from the reference and travels 0.0336 in; this point is halfway to that edge.
From Work and Energy
1. Steady lift in SI: A hoist raises a 320 kg load 4.5 m at a steady speed. How much work is done against gravity? Use standard gravity, g0 = 9.80665 m/s2.
Show worked solution
- The load is a mass, so turn it into the force needed to lift it:
F = mg = 320 kg × 9.80665 m/s2 = 3,138.128 N. W = Fs = 3,138.128 N × 4.5 m = 14,121.576 J.
The work done is 14.1 kJ. Two assumptions are doing real work in that answer: the lift is steady, so no extra force goes into accelerating the load, and the force acts along the direction of travel for the whole 4.5 m.
From Work and Energy
2. A lift in foot pounds-force: How much work is required to lift a 750 lb load 22 ft? Give the answer in joules.
Show worked solution
- The force needed to lift steadily is the load’s weight, 750 lbf, acting along the 22 ft of travel:
W = 750 lbf × 22 ft = 16,500 ft·lbf.
Then convert:
16,500 ft·lbf × = 22,370.997 J
or 22.4 kJ. Note the word order in the working unit: energy is ft·lbf, and it converts to the joule. The 750 lb figure is treated as a weight in pound-force, which is what the formula needs — not a mass in pound-mass.
From Work and Energy
3. A meter reading in megajoules: An electricity meter records 8.4 kW·h over one shift. How many megajoules of energy is that?
Show worked solution
A kilowatt hour is a power multiplied by a time and nothing more: one kilowatt sustained for one hour. So 1 kW·h = 1,000 W × 3,600 s = 3,600,000 J exactly, which is 3.6 MJ.
8.4 kW·h × = 30.24 MJ
The shift used 30.24 MJ. The kilowatt hour is not itself an SI unit — NIST lists it among the units to be converted into SI — but the factor is exact, so nothing is lost in the conversion.
From Work and Energy
4. Work from a torque: A fastener is run down with a steady 45 N·m through 2.5 revolutions. How much work is done, in joules?
Show worked solution
- The angle has to be in radians before it can multiply a torque:
2.5 r × 6.283185 rad per r = 15.70796 rad. W = τθ = 45 N·m × 15.70796 rad = 706.858 J.
The work done is 707 J. Nothing was converted between N·m and J along the way, and nothing needed to be: the radian is a length over a length, value 1, so multiplying by an angle in radians leaves the units alone. Feed the same formula 2.5 revolutions or 900 degrees directly and the answer is wrong, because both of those carry a chosen count inside them.
The steady 45 N·m is an idealization — a real run-down torque climbs as the joint seats.
From Power and Efficiency
5. Horsepower for a lift: A 750 lb load is raised 22 ft in 9 s. How much power does that take, in hp?
Show worked solution
W = 750 lbf × 22 ft = 16,500 ft·lbf.P = W/t = 16,500 ft·lbf ÷ 9 s = 1,833.33 ft·lbf/s.- One horsepower is defined as 550 ft·lbf/s, so dividing by 550 is a unit conversion, not a fudge:
1,833.33 ÷ 550 = 3.3333 hp.
The lift needs 3.33 hp delivered to the load. A motor for the job needs more than that, because no drive is perfectly efficient. Cross-check in SI if you like: 16,500 ft·lbf × 1.355818 = 22,371 J, and 22,371 J ÷ 9 s = 2,486 W, which is 2,486 ÷ 745.6999 = 3.33 hp.
From Power and Efficiency
6. Power from torque and speed: A motor delivers 12 N·m at 1,750 r/min. What is its mechanical power output, in W?
Show worked solution
P = τωneeds the speed in rad/s:1,750 r/min × 0.1047198 rad/s per r/min = 183.2596 rad/s.P = 12 N·m × 183.2596 rad/s = 2,199.1 W.
The output is 2,199 W, or about 2.20 kW. Substituting 1,750 straight into the formula gives 21,000 — wrong by the factor of about 9.55 that separates r/min from rad/s. In coherent SI units the relationship needs no constant at all, which is exactly why the speed has to arrive in rad/s.
From Power and Efficiency
7. The torque-and-rpm shortcut: A drive delivers 250 lbf·in at 1,200 r/min. What is the power, in hp?
Show worked solution
The shortcut takes the torque in pound-force inches and the speed in r/min:
hp = (250 lbf·in × 1,200 r/min) ÷ 63,025 = 300,000 ÷ 63,025 = 4.760 hp
The answer is 4.76 hp. Check it the long way, since the constant is only a folded-up set of conversions: 250 lbf·in = 20.833 lbf·ft, 1,200 r/min = 125.664 rad/s, so P = 20.833 × 125.664 = 2,618.0 ft·lbf/s, and 2,618.0 ÷ 550 = 4.760 hp. The two routes agree to four figures.
Give the shortcut a torque in lbf·ft instead and the answer comes out twelve times too small — the constant already contains the inch-to-foot step.
From Power and Efficiency
8. The volts-and-amps shortcut: A direct-current motor draws 14 A at 24 V. What is that electrical input power, in hp?
Show worked solution
- Electrical power first:
P = VI = 24 V × 14 A = 336 W. - Then convert:
336 W ÷ 746 W per hp = 0.4504 hp.
The input is 0.450 hp. The 746 is exact because NIST defines a separate unit, horsepower (electric), as 746 W.
The condition sits in step 1, and this problem satisfies it: volts times amps gives real power in watts on direct current or a purely resistive load. On alternating current it gives apparent power in volt-amperes, and the real power is that multiplied by the power factor — so the same shortcut used on an AC motor without its power factor reads high. This is also the motor’s input; its mechanical output is less.
From Power and Efficiency
9. Hoist efficiency: A hoist motor draws 2.2 kW and raises a 900 N load at a steady 2.0 m/s. What is the efficiency of the system, as a percentage?
Show worked solution
- Useful output, using
P = Fv:900 N × 2.0 m/s = 1,800 W. η = Pout/Pin = 1,800 W ÷ 2,200 W = 0.8182.
The system is 81.8% efficient. The ratio is two like quantities, so the watts cancel and the answer is a plain number. The missing 400 W has not vanished — it is heat in the windings, the gearbox, and the bearings. An answer above 100% would mean an error in the numbers, not a remarkable machine.
This is a system efficiency, measured from the electrical input to the load. The motor alone would quote a higher figure, over a different boundary.